How do you find #dy/dx# given #x=3y^(1/3)+2y#?

Now proceeding my differentiating:

Factoring:

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Differentiating x = 3y^(1/3) + 2To find ( \frac{dy}{dx} ) given ( x = 3y^{1/3} +To find dy/dx, we differentiate the given equation implicitly with respect to x.

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Differentiating x = 3y^(1/3) + 2y with respect to x, we get:

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Differentiating x = 3y^(1/3) + 2y with respect to x, we get:

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Differentiating x = 3y^(1/3) + 2y with respect to x, we get:

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Differentiating x = 3y^(1/3) + 2y with respect to x, we get:

1 = (To find ( \frac{dy}{dx} ) given ( x = 3y^{1/3} + 2y ), differentiate both sides of the equation with respect to ( x )To find dy/dx, we differentiate the given equation implicitly with respect to x.

Differentiating x = 3y^(1/3) + 2y with respect to x, we get:

1 = (1To find ( \frac{dy}{dx} ) given ( x = 3y^{1/3} + 2y ), differentiate both sides of the equation with respect to ( x ) usingTo find dy/dx, we differentiate the given equation implicitly with respect to x.

Differentiating x = 3y^(1/3) + 2y with respect to x, we get:

1 = (1/To find ( \frac{dy}{dx} ) given ( x = 3y^{1/3} + 2y ), differentiate both sides of the equation with respect to ( x ) using implicit differentiationTo find dy/dx, we differentiate the given equation implicitly with respect to x.

Differentiating x = 3y^(1/3) + 2y with respect to x, we get:

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Differentiating x = 3y^(1/3) + 2y with respect to x, we get:

1 = (1/3)To find ( \frac{dy}{dx} ) given ( x = 3y^{1/3} + 2y ), differentiate both sides of the equation with respect to ( x ) using implicit differentiation.

To find dy/dx, we differentiate the given equation implicitly with respect to x.

Differentiating x = 3y^(1/3) + 2y with respect to x, we get:

1 = (1/3) *To find ( \frac{dy}{dx} ) given ( x = 3y^{1/3} + 2y ), differentiate both sides of the equation with respect to ( x ) using implicit differentiation.

DifferentTo find dy/dx, we differentiate the given equation implicitly with respect to x.

Differentiating x = 3y^(1/3) + 2y with respect to x, we get:

1 = (1/3) * To find ( \frac{dy}{dx} ) given ( x = 3y^{1/3} + 2y ), differentiate both sides of the equation with respect to ( x ) using implicit differentiation.

Differentiating (To find dy/dx, we differentiate the given equation implicitly with respect to x.

Differentiating x = 3y^(1/3) + 2y with respect to x, we get:

1 = (1/3) * 3To find ( \frac{dy}{dx} ) given ( x = 3y^{1/3} + 2y ), differentiate both sides of the equation with respect to ( x ) using implicit differentiation.

Differentiating ( xTo find dy/dx, we differentiate the given equation implicitly with respect to x.

Differentiating x = 3y^(1/3) + 2y with respect to x, we get:

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Differentiating ( x )To find dy/dx, we differentiate the given equation implicitly with respect to x.

Differentiating x = 3y^(1/3) + 2y with respect to x, we get:

1 = (1/3) * 3 * yTo find ( \frac{dy}{dx} ) given ( x = 3y^{1/3} + 2y ), differentiate both sides of the equation with respect to ( x ) using implicit differentiation.

Differentiating ( x ) withTo find dy/dx, we differentiate the given equation implicitly with respect to x.

Differentiating x = 3y^(1/3) + 2y with respect to x, we get:

1 = (1/3) * 3 * y^To find ( \frac{dy}{dx} ) given ( x = 3y^{1/3} + 2y ), differentiate both sides of the equation with respect to ( x ) using implicit differentiation.

Differentiating ( x ) with respectTo find dy/dx, we differentiate the given equation implicitly with respect to x.

Differentiating x = 3y^(1/3) + 2y with respect to x, we get:

1 = (1/3) * 3 * y^(-To find ( \frac{dy}{dx} ) given ( x = 3y^{1/3} + 2y ), differentiate both sides of the equation with respect to ( x ) using implicit differentiation.

Differentiating ( x ) with respect toTo find dy/dx, we differentiate the given equation implicitly with respect to x.

Differentiating x = 3y^(1/3) + 2y with respect to x, we get:

1 = (1/3) * 3 * y^(-2To find ( \frac{dy}{dx} ) given ( x = 3y^{1/3} + 2y ), differentiate both sides of the equation with respect to ( x ) using implicit differentiation.

Differentiating ( x ) with respect to (To find dy/dx, we differentiate the given equation implicitly with respect to x.

Differentiating x = 3y^(1/3) + 2y with respect to x, we get:

1 = (1/3) * 3 * y^(-2/To find ( \frac{dy}{dx} ) given ( x = 3y^{1/3} + 2y ), differentiate both sides of the equation with respect to ( x ) using implicit differentiation.

Differentiating ( x ) with respect to ( xTo find dy/dx, we differentiate the given equation implicitly with respect to x.

Differentiating x = 3y^(1/3) + 2y with respect to x, we get:

1 = (1/3) * 3 * y^(-2/3To find ( \frac{dy}{dx} ) given ( x = 3y^{1/3} + 2y ), differentiate both sides of the equation with respect to ( x ) using implicit differentiation.

Differentiating ( x ) with respect to ( x \To find dy/dx, we differentiate the given equation implicitly with respect to x.

Differentiating x = 3y^(1/3) + 2y with respect to x, we get:

1 = (1/3) * 3 * y^(-2/3)To find ( \frac{dy}{dx} ) given ( x = 3y^{1/3} + 2y ), differentiate both sides of the equation with respect to ( x ) using implicit differentiation.

Differentiating ( x ) with respect to ( x ) gives (To find dy/dx, we differentiate the given equation implicitly with respect to x.

Differentiating x = 3y^(1/3) + 2y with respect to x, we get:

1 = (1/3) * 3 * y^(-2/3) * dy/dTo find ( \frac{dy}{dx} ) given ( x = 3y^{1/3} + 2y ), differentiate both sides of the equation with respect to ( x ) using implicit differentiation.

Differentiating ( x ) with respect to ( x ) gives ( To find dy/dx, we differentiate the given equation implicitly with respect to x.

Differentiating x = 3y^(1/3) + 2y with respect to x, we get:

1 = (1/3) * 3 * y^(-2/3) * dy/dxTo find ( \frac{dy}{dx} ) given ( x = 3y^{1/3} + 2y ), differentiate both sides of the equation with respect to ( x ) using implicit differentiation.

Differentiating ( x ) with respect to ( x ) gives ( 1To find dy/dx, we differentiate the given equation implicitly with respect to x.

Differentiating x = 3y^(1/3) + 2y with respect to x, we get:

1 = (1/3) * 3 * y^(-2/3) * dy/dx +To find ( \frac{dy}{dx} ) given ( x = 3y^{1/3} + 2y ), differentiate both sides of the equation with respect to ( x ) using implicit differentiation.

Differentiating ( x ) with respect to ( x ) gives ( 1 \To find dy/dx, we differentiate the given equation implicitly with respect to x.

Differentiating x = 3y^(1/3) + 2y with respect to x, we get:

1 = (1/3) * 3 * y^(-2/3) * dy/dx + To find ( \frac{dy}{dx} ) given ( x = 3y^{1/3} + 2y ), differentiate both sides of the equation with respect to ( x ) using implicit differentiation.

Differentiating ( x ) with respect to ( x ) gives ( 1 ). To find dy/dx, we differentiate the given equation implicitly with respect to x.

Differentiating x = 3y^(1/3) + 2y with respect to x, we get:

1 = (1/3) * 3 * y^(-2/3) * dy/dx + 2To find ( \frac{dy}{dx} ) given ( x = 3y^{1/3} + 2y ), differentiate both sides of the equation with respect to ( x ) using implicit differentiation.

Differentiating ( x ) with respect to ( x ) gives ( 1 ). DifferentiatingTo find dy/dx, we differentiate the given equation implicitly with respect to x.

Differentiating x = 3y^(1/3) + 2y with respect to x, we get:

1 = (1/3) * 3 * y^(-2/3) * dy/dx + 2 *To find ( \frac{dy}{dx} ) given ( x = 3y^{1/3} + 2y ), differentiate both sides of the equation with respect to ( x ) using implicit differentiation.

Differentiating ( x ) with respect to ( x ) gives ( 1 ). Differentiating (To find dy/dx, we differentiate the given equation implicitly with respect to x.

Differentiating x = 3y^(1/3) + 2y with respect to x, we get:

1 = (1/3) * 3 * y^(-2/3) * dy/dx + 2 * dyTo find ( \frac{dy}{dx} ) given ( x = 3y^{1/3} + 2y ), differentiate both sides of the equation with respect to ( x ) using implicit differentiation.

Differentiating ( x ) with respect to ( x ) gives ( 1 ). Differentiating ( 3y^{1/3} )To find dy/dx, we differentiate the given equation implicitly with respect to x.

Differentiating x = 3y^(1/3) + 2y with respect to x, we get:

1 = (1/3) * 3 * y^(-2/3) * dy/dx + 2 * dy/dx

Simplifying this,To find ( \frac{dy}{dx} ) given ( x = 3y^{1/3} + 2y ), differentiate both sides of the equation with respect to ( x ) using implicit differentiation.

Differentiating ( x ) with respect to ( x ) gives ( 1 ). Differentiating ( 3y^{1/3} ) withTo find dy/dx, we differentiate the given equation implicitly with respect to x.

Differentiating x = 3y^(1/3) + 2y with respect to x, we get:

1 = (1/3) * 3 * y^(-2/3) * dy/dx + 2 * dy/dx

Simplifying this, weTo find ( \frac{dy}{dx} ) given ( x = 3y^{1/3} + 2y ), differentiate both sides of the equation with respect to ( x ) using implicit differentiation.

Differentiating ( x ) with respect to ( x ) gives ( 1 ). Differentiating ( 3y^{1/3} ) with respectTo find dy/dx, we differentiate the given equation implicitly with respect to x.

Differentiating x = 3y^(1/3) + 2y with respect to x, we get:

1 = (1/3) * 3 * y^(-2/3) * dy/dx + 2 * dy/dx

Simplifying this, we haveTo find ( \frac{dy}{dx} ) given ( x = 3y^{1/3} + 2y ), differentiate both sides of the equation with respect to ( x ) using implicit differentiation.

Differentiating ( x ) with respect to ( x ) gives ( 1 ). Differentiating ( 3y^{1/3} ) with respect toTo find dy/dx, we differentiate the given equation implicitly with respect to x.

Differentiating x = 3y^(1/3) + 2y with respect to x, we get:

1 = (1/3) * 3 * y^(-2/3) * dy/dx + 2 * dy/dx

Simplifying this, we have:

To find ( \frac{dy}{dx} ) given ( x = 3y^{1/3} + 2y ), differentiate both sides of the equation with respect to ( x ) using implicit differentiation.

Differentiating ( x ) with respect to ( x ) gives ( 1 ). Differentiating ( 3y^{1/3} ) with respect to (To find dy/dx, we differentiate the given equation implicitly with respect to x.

Differentiating x = 3y^(1/3) + 2y with respect to x, we get:

1 = (1/3) * 3 * y^(-2/3) * dy/dx + 2 * dy/dx

Simplifying this, we have:

1To find ( \frac{dy}{dx} ) given ( x = 3y^{1/3} + 2y ), differentiate both sides of the equation with respect to ( x ) using implicit differentiation.

Differentiating ( x ) with respect to ( x ) gives ( 1 ). Differentiating ( 3y^{1/3} ) with respect to ( x \To find dy/dx, we differentiate the given equation implicitly with respect to x.

Differentiating x = 3y^(1/3) + 2y with respect to x, we get:

1 = (1/3) * 3 * y^(-2/3) * dy/dx + 2 * dy/dx

Simplifying this, we have:

1 =To find ( \frac{dy}{dx} ) given ( x = 3y^{1/3} + 2y ), differentiate both sides of the equation with respect to ( x ) using implicit differentiation.

Differentiating ( x ) with respect to ( x ) gives ( 1 ). Differentiating ( 3y^{1/3} ) with respect to ( x )To find dy/dx, we differentiate the given equation implicitly with respect to x.

Differentiating x = 3y^(1/3) + 2y with respect to x, we get:

1 = (1/3) * 3 * y^(-2/3) * dy/dx + 2 * dy/dx

Simplifying this, we have:

1 = yTo find ( \frac{dy}{dx} ) given ( x = 3y^{1/3} + 2y ), differentiate both sides of the equation with respect to ( x ) using implicit differentiation.

Differentiating ( x ) with respect to ( x ) gives ( 1 ). Differentiating ( 3y^{1/3} ) with respect to ( x ) using theTo find dy/dx, we differentiate the given equation implicitly with respect to x.

Differentiating x = 3y^(1/3) + 2y with respect to x, we get:

1 = (1/3) * 3 * y^(-2/3) * dy/dx + 2 * dy/dx

Simplifying this, we have:

1 = y^To find ( \frac{dy}{dx} ) given ( x = 3y^{1/3} + 2y ), differentiate both sides of the equation with respect to ( x ) using implicit differentiation.

Differentiating ( x ) with respect to ( x ) gives ( 1 ). Differentiating ( 3y^{1/3} ) with respect to ( x ) using the chainTo find dy/dx, we differentiate the given equation implicitly with respect to x.

Differentiating x = 3y^(1/3) + 2y with respect to x, we get:

1 = (1/3) * 3 * y^(-2/3) * dy/dx + 2 * dy/dx

Simplifying this, we have:

1 = y^(-To find ( \frac{dy}{dx} ) given ( x = 3y^{1/3} + 2y ), differentiate both sides of the equation with respect to ( x ) using implicit differentiation.

Differentiating ( x ) with respect to ( x ) gives ( 1 ). Differentiating ( 3y^{1/3} ) with respect to ( x ) using the chain ruleTo find dy/dx, we differentiate the given equation implicitly with respect to x.

Differentiating x = 3y^(1/3) + 2y with respect to x, we get:

1 = (1/3) * 3 * y^(-2/3) * dy/dx + 2 * dy/dx

Simplifying this, we have:

1 = y^(-2To find ( \frac{dy}{dx} ) given ( x = 3y^{1/3} + 2y ), differentiate both sides of the equation with respect to ( x ) using implicit differentiation.

Differentiating ( x ) with respect to ( x ) gives ( 1 ). Differentiating ( 3y^{1/3} ) with respect to ( x ) using the chain rule givesTo find dy/dx, we differentiate the given equation implicitly with respect to x.

Differentiating x = 3y^(1/3) + 2y with respect to x, we get:

1 = (1/3) * 3 * y^(-2/3) * dy/dx + 2 * dy/dx

Simplifying this, we have:

1 = y^(-2/To find ( \frac{dy}{dx} ) given ( x = 3y^{1/3} + 2y ), differentiate both sides of the equation with respect to ( x ) using implicit differentiation.

Differentiating ( x ) with respect to ( x ) gives ( 1 ). Differentiating ( 3y^{1/3} ) with respect to ( x ) using the chain rule gives (To find dy/dx, we differentiate the given equation implicitly with respect to x.

Differentiating x = 3y^(1/3) + 2y with respect to x, we get:

1 = (1/3) * 3 * y^(-2/3) * dy/dx + 2 * dy/dx

Simplifying this, we have:

1 = y^(-2/3To find ( \frac{dy}{dx} ) given ( x = 3y^{1/3} + 2y ), differentiate both sides of the equation with respect to ( x ) using implicit differentiation.

Differentiating ( x ) with respect to ( x ) gives ( 1 ). Differentiating ( 3y^{1/3} ) with respect to ( x ) using the chain rule gives ( \To find dy/dx, we differentiate the given equation implicitly with respect to x.

Differentiating x = 3y^(1/3) + 2y with respect to x, we get:

1 = (1/3) * 3 * y^(-2/3) * dy/dx + 2 * dy/dx

Simplifying this, we have:

1 = y^(-2/3)To find ( \frac{dy}{dx} ) given ( x = 3y^{1/3} + 2y ), differentiate both sides of the equation with respect to ( x ) using implicit differentiation.

Differentiating ( x ) with respect to ( x ) gives ( 1 ). Differentiating ( 3y^{1/3} ) with respect to ( x ) using the chain rule gives ( \frac{To find dy/dx, we differentiate the given equation implicitly with respect to x.

Differentiating x = 3y^(1/3) + 2y with respect to x, we get:

1 = (1/3) * 3 * y^(-2/3) * dy/dx + 2 * dy/dx

Simplifying this, we have:

1 = y^(-2/3) *To find ( \frac{dy}{dx} ) given ( x = 3y^{1/3} + 2y ), differentiate both sides of the equation with respect to ( x ) using implicit differentiation.

Differentiating ( x ) with respect to ( x ) gives ( 1 ). Differentiating ( 3y^{1/3} ) with respect to ( x ) using the chain rule gives ( \frac{1To find dy/dx, we differentiate the given equation implicitly with respect to x.

Differentiating x = 3y^(1/3) + 2y with respect to x, we get:

1 = (1/3) * 3 * y^(-2/3) * dy/dx + 2 * dy/dx

Simplifying this, we have:

1 = y^(-2/3) * dyTo find ( \frac{dy}{dx} ) given ( x = 3y^{1/3} + 2y ), differentiate both sides of the equation with respect to ( x ) using implicit differentiation.

Differentiating ( x ) with respect to ( x ) gives ( 1 ). Differentiating ( 3y^{1/3} ) with respect to ( x ) using the chain rule gives ( \frac{1}{3To find dy/dx, we differentiate the given equation implicitly with respect to x.

Differentiating x = 3y^(1/3) + 2y with respect to x, we get:

1 = (1/3) * 3 * y^(-2/3) * dy/dx + 2 * dy/dx

Simplifying this, we have:

1 = y^(-2/3) * dy/dTo find ( \frac{dy}{dx} ) given ( x = 3y^{1/3} + 2y ), differentiate both sides of the equation with respect to ( x ) using implicit differentiation.

Differentiating ( x ) with respect to ( x ) gives ( 1 ). Differentiating ( 3y^{1/3} ) with respect to ( x ) using the chain rule gives ( \frac{1}{3}To find dy/dx, we differentiate the given equation implicitly with respect to x.

Differentiating x = 3y^(1/3) + 2y with respect to x, we get:

1 = (1/3) * 3 * y^(-2/3) * dy/dx + 2 * dy/dx

Simplifying this, we have:

1 = y^(-2/3) * dy/dxTo find ( \frac{dy}{dx} ) given ( x = 3y^{1/3} + 2y ), differentiate both sides of the equation with respect to ( x ) using implicit differentiation.

Differentiating ( x ) with respect to ( x ) gives ( 1 ). Differentiating ( 3y^{1/3} ) with respect to ( x ) using the chain rule gives ( \frac{1}{3} \To find dy/dx, we differentiate the given equation implicitly with respect to x.

Differentiating x = 3y^(1/3) + 2y with respect to x, we get:

1 = (1/3) * 3 * y^(-2/3) * dy/dx + 2 * dy/dx

Simplifying this, we have:

1 = y^(-2/3) * dy/dx +To find ( \frac{dy}{dx} ) given ( x = 3y^{1/3} + 2y ), differentiate both sides of the equation with respect to ( x ) using implicit differentiation.

Differentiating ( x ) with respect to ( x ) gives ( 1 ). Differentiating ( 3y^{1/3} ) with respect to ( x ) using the chain rule gives ( \frac{1}{3} \cdot 3To find dy/dx, we differentiate the given equation implicitly with respect to x.

Differentiating x = 3y^(1/3) + 2y with respect to x, we get:

1 = (1/3) * 3 * y^(-2/3) * dy/dx + 2 * dy/dx

Simplifying this, we have:

1 = y^(-2/3) * dy/dx + To find ( \frac{dy}{dx} ) given ( x = 3y^{1/3} + 2y ), differentiate both sides of the equation with respect to ( x ) using implicit differentiation.

Differentiating ( x ) with respect to ( x ) gives ( 1 ). Differentiating ( 3y^{1/3} ) with respect to ( x ) using the chain rule gives ( \frac{1}{3} \cdot 3 \To find dy/dx, we differentiate the given equation implicitly with respect to x.

Differentiating x = 3y^(1/3) + 2y with respect to x, we get:

1 = (1/3) * 3 * y^(-2/3) * dy/dx + 2 * dy/dx

Simplifying this, we have:

1 = y^(-2/3) * dy/dx + 2To find ( \frac{dy}{dx} ) given ( x = 3y^{1/3} + 2y ), differentiate both sides of the equation with respect to ( x ) using implicit differentiation.

Differentiating ( x ) with respect to ( x ) gives ( 1 ). Differentiating ( 3y^{1/3} ) with respect to ( x ) using the chain rule gives ( \frac{1}{3} \cdot 3 \cdot yTo find dy/dx, we differentiate the given equation implicitly with respect to x.

Differentiating x = 3y^(1/3) + 2y with respect to x, we get:

1 = (1/3) * 3 * y^(-2/3) * dy/dx + 2 * dy/dx

Simplifying this, we have:

1 = y^(-2/3) * dy/dx + 2 *To find ( \frac{dy}{dx} ) given ( x = 3y^{1/3} + 2y ), differentiate both sides of the equation with respect to ( x ) using implicit differentiation.

Differentiating ( x ) with respect to ( x ) gives ( 1 ). Differentiating ( 3y^{1/3} ) with respect to ( x ) using the chain rule gives ( \frac{1}{3} \cdot 3 \cdot y^{-To find dy/dx, we differentiate the given equation implicitly with respect to x.

Differentiating x = 3y^(1/3) + 2y with respect to x, we get:

1 = (1/3) * 3 * y^(-2/3) * dy/dx + 2 * dy/dx

Simplifying this, we have:

1 = y^(-2/3) * dy/dx + 2 * dyTo find ( \frac{dy}{dx} ) given ( x = 3y^{1/3} + 2y ), differentiate both sides of the equation with respect to ( x ) using implicit differentiation.

Differentiating ( x ) with respect to ( x ) gives ( 1 ). Differentiating ( 3y^{1/3} ) with respect to ( x ) using the chain rule gives ( \frac{1}{3} \cdot 3 \cdot y^{-2/3} \cdot \fracTo find dy/dx, we differentiate the given equation implicitly with respect to x.

Differentiating x = 3y^(1/3) + 2y with respect to x, we get:

1 = (1/3) * 3 * y^(-2/3) * dy/dx + 2 * dy/dx

Simplifying this, we have:

1 = y^(-2/3) * dy/dx + 2 * dy/dx

Multiplying throughoutTo find ( \frac{dy}{dx} ) given ( x = 3y^{1/3} + 2y ), differentiate both sides of the equation with respect to ( x ) using implicit differentiation.

Differentiating ( x ) with respect to ( x ) gives ( 1 ). Differentiating ( 3y^{1/3} ) with respect to ( x ) using the chain rule gives ( \frac{1}{3} \cdot 3 \cdot y^{-2/3} \cdot \frac{To find dy/dx, we differentiate the given equation implicitly with respect to x.

Differentiating x = 3y^(1/3) + 2y with respect to x, we get:

1 = (1/3) * 3 * y^(-2/3) * dy/dx + 2 * dy/dx

Simplifying this, we have:

1 = y^(-2/3) * dy/dx + 2 * dy/dx

Multiplying throughout by y^(2To find ( \frac{dy}{dx} ) given ( x = 3y^{1/3} + 2y ), differentiate both sides of the equation with respect to ( x ) using implicit differentiation.

Differentiating ( x ) with respect to ( x ) gives ( 1 ). Differentiating ( 3y^{1/3} ) with respect to ( x ) using the chain rule gives ( \frac{1}{3} \cdot 3 \cdot y^{-2/3} \cdot \frac{dy}{To find dy/dx, we differentiate the given equation implicitly with respect to x.

Differentiating x = 3y^(1/3) + 2y with respect to x, we get:

1 = (1/3) * 3 * y^(-2/3) * dy/dx + 2 * dy/dx

Simplifying this, we have:

1 = y^(-2/3) * dy/dx + 2 * dy/dx

Multiplying throughout by y^(2/To find ( \frac{dy}{dx} ) given ( x = 3y^{1/3} + 2y ), differentiate both sides of the equation with respect to ( x ) using implicit differentiation.

Differentiating ( x ) with respect to ( x ) gives ( 1 ). Differentiating ( 3y^{1/3} ) with respect to ( x ) using the chain rule gives ( \frac{1}{3} \cdot 3 \cdot y^{-2/3} \cdot \frac{dy}{dxTo find dy/dx, we differentiate the given equation implicitly with respect to x.

Differentiating x = 3y^(1/3) + 2y with respect to x, we get:

1 = (1/3) * 3 * y^(-2/3) * dy/dx + 2 * dy/dx

Simplifying this, we have:

1 = y^(-2/3) * dy/dx + 2 * dy/dx

Multiplying throughout by y^(2/3To find ( \frac{dy}{dx} ) given ( x = 3y^{1/3} + 2y ), differentiate both sides of the equation with respect to ( x ) using implicit differentiation.

Differentiating ( x ) with respect to ( x ) gives ( 1 ). Differentiating ( 3y^{1/3} ) with respect to ( x ) using the chain rule gives ( \frac{1}{3} \cdot 3 \cdot y^{-2/3} \cdot \frac{dy}{dx}To find dy/dx, we differentiate the given equation implicitly with respect to x.

Differentiating x = 3y^(1/3) + 2y with respect to x, we get:

1 = (1/3) * 3 * y^(-2/3) * dy/dx + 2 * dy/dx

Simplifying this, we have:

1 = y^(-2/3) * dy/dx + 2 * dy/dx

Multiplying throughout by y^(2/3),To find ( \frac{dy}{dx} ) given ( x = 3y^{1/3} + 2y ), differentiate both sides of the equation with respect to ( x ) using implicit differentiation.

Differentiating ( x ) with respect to ( x ) gives ( 1 ). Differentiating ( 3y^{1/3} ) with respect to ( x ) using the chain rule gives ( \frac{1}{3} \cdot 3 \cdot y^{-2/3} \cdot \frac{dy}{dx} =To find dy/dx, we differentiate the given equation implicitly with respect to x.

Differentiating x = 3y^(1/3) + 2y with respect to x, we get:

1 = (1/3) * 3 * y^(-2/3) * dy/dx + 2 * dy/dx

Simplifying this, we have:

1 = y^(-2/3) * dy/dx + 2 * dy/dx

Multiplying throughout by y^(2/3), weTo find ( \frac{dy}{dx} ) given ( x = 3y^{1/3} + 2y ), differentiate both sides of the equation with respect to ( x ) using implicit differentiation.

Differentiating ( x ) with respect to ( x ) gives ( 1 ). Differentiating ( 3y^{1/3} ) with respect to ( x ) using the chain rule gives ( \frac{1}{3} \cdot 3 \cdot y^{-2/3} \cdot \frac{dy}{dx} = yTo find dy/dx, we differentiate the given equation implicitly with respect to x.

Differentiating x = 3y^(1/3) + 2y with respect to x, we get:

1 = (1/3) * 3 * y^(-2/3) * dy/dx + 2 * dy/dx

Simplifying this, we have:

1 = y^(-2/3) * dy/dx + 2 * dy/dx

Multiplying throughout by y^(2/3), we getTo find ( \frac{dy}{dx} ) given ( x = 3y^{1/3} + 2y ), differentiate both sides of the equation with respect to ( x ) using implicit differentiation.

Differentiating ( x ) with respect to ( x ) gives ( 1 ). Differentiating ( 3y^{1/3} ) with respect to ( x ) using the chain rule gives ( \frac{1}{3} \cdot 3 \cdot y^{-2/3} \cdot \frac{dy}{dx} = y^{-To find dy/dx, we differentiate the given equation implicitly with respect to x.

Differentiating x = 3y^(1/3) + 2y with respect to x, we get:

1 = (1/3) * 3 * y^(-2/3) * dy/dx + 2 * dy/dx

Simplifying this, we have:

1 = y^(-2/3) * dy/dx + 2 * dy/dx

Multiplying throughout by y^(2/3), we get:

To find ( \frac{dy}{dx} ) given ( x = 3y^{1/3} + 2y ), differentiate both sides of the equation with respect to ( x ) using implicit differentiation.

Differentiating ( x ) with respect to ( x ) gives ( 1 ). Differentiating ( 3y^{1/3} ) with respect to ( x ) using the chain rule gives ( \frac{1}{3} \cdot 3 \cdot y^{-2/3} \cdot \frac{dy}{dx} = y^{-2To find dy/dx, we differentiate the given equation implicitly with respect to x.

Differentiating x = 3y^(1/3) + 2y with respect to x, we get:

1 = (1/3) * 3 * y^(-2/3) * dy/dx + 2 * dy/dx

Simplifying this, we have:

1 = y^(-2/3) * dy/dx + 2 * dy/dx

Multiplying throughout by y^(2/3), we get:

yTo find ( \frac{dy}{dx} ) given ( x = 3y^{1/3} + 2y ), differentiate both sides of the equation with respect to ( x ) using implicit differentiation.

Differentiating ( x ) with respect to ( x ) gives ( 1 ). Differentiating ( 3y^{1/3} ) with respect to ( x ) using the chain rule gives ( \frac{1}{3} \cdot 3 \cdot y^{-2/3} \cdot \frac{dy}{dx} = y^{-2/To find dy/dx, we differentiate the given equation implicitly with respect to x.

Differentiating x = 3y^(1/3) + 2y with respect to x, we get:

1 = (1/3) * 3 * y^(-2/3) * dy/dx + 2 * dy/dx

Simplifying this, we have:

1 = y^(-2/3) * dy/dx + 2 * dy/dx

Multiplying throughout by y^(2/3), we get:

y^(To find ( \frac{dy}{dx} ) given ( x = 3y^{1/3} + 2y ), differentiate both sides of the equation with respect to ( x ) using implicit differentiation.

Differentiating ( x ) with respect to ( x ) gives ( 1 ). Differentiating ( 3y^{1/3} ) with respect to ( x ) using the chain rule gives ( \frac{1}{3} \cdot 3 \cdot y^{-2/3} \cdot \frac{dy}{dx} = y^{-2/3To find dy/dx, we differentiate the given equation implicitly with respect to x.

Differentiating x = 3y^(1/3) + 2y with respect to x, we get:

1 = (1/3) * 3 * y^(-2/3) * dy/dx + 2 * dy/dx

Simplifying this, we have:

1 = y^(-2/3) * dy/dx + 2 * dy/dx

Multiplying throughout by y^(2/3), we get:

y^(2/To find ( \frac{dy}{dx} ) given ( x = 3y^{1/3} + 2y ), differentiate both sides of the equation with respect to ( x ) using implicit differentiation.

Differentiating ( x ) with respect to ( x ) gives ( 1 ). Differentiating ( 3y^{1/3} ) with respect to ( x ) using the chain rule gives ( \frac{1}{3} \cdot 3 \cdot y^{-2/3} \cdot \frac{dy}{dx} = y^{-2/3}To find dy/dx, we differentiate the given equation implicitly with respect to x.

Differentiating x = 3y^(1/3) + 2y with respect to x, we get:

1 = (1/3) * 3 * y^(-2/3) * dy/dx + 2 * dy/dx

Simplifying this, we have:

1 = y^(-2/3) * dy/dx + 2 * dy/dx

Multiplying throughout by y^(2/3), we get:

y^(2/3To find ( \frac{dy}{dx} ) given ( x = 3y^{1/3} + 2y ), differentiate both sides of the equation with respect to ( x ) using implicit differentiation.

Differentiating ( x ) with respect to ( x ) gives ( 1 ). Differentiating ( 3y^{1/3} ) with respect to ( x ) using the chain rule gives ( \frac{1}{3} \cdot 3 \cdot y^{-2/3} \cdot \frac{dy}{dx} = y^{-2/3} \To find dy/dx, we differentiate the given equation implicitly with respect to x.

Differentiating x = 3y^(1/3) + 2y with respect to x, we get:

1 = (1/3) * 3 * y^(-2/3) * dy/dx + 2 * dy/dx

Simplifying this, we have:

1 = y^(-2/3) * dy/dx + 2 * dy/dx

Multiplying throughout by y^(2/3), we get:

y^(2/3) =To find ( \frac{dy}{dx} ) given ( x = 3y^{1/3} + 2y ), differentiate both sides of the equation with respect to ( x ) using implicit differentiation.

Differentiating ( x ) with respect to ( x ) gives ( 1 ). Differentiating ( 3y^{1/3} ) with respect to ( x ) using the chain rule gives ( \frac{1}{3} \cdot 3 \cdot y^{-2/3} \cdot \frac{dy}{dx} = y^{-2/3} \cdotTo find dy/dx, we differentiate the given equation implicitly with respect to x.

Differentiating x = 3y^(1/3) + 2y with respect to x, we get:

1 = (1/3) * 3 * y^(-2/3) * dy/dx + 2 * dy/dx

Simplifying this, we have:

1 = y^(-2/3) * dy/dx + 2 * dy/dx

Multiplying throughout by y^(2/3), we get:

y^(2/3) = dyTo find ( \frac{dy}{dx} ) given ( x = 3y^{1/3} + 2y ), differentiate both sides of the equation with respect to ( x ) using implicit differentiation.

Differentiating ( x ) with respect to ( x ) gives ( 1 ). Differentiating ( 3y^{1/3} ) with respect to ( x ) using the chain rule gives ( \frac{1}{3} \cdot 3 \cdot y^{-2/3} \cdot \frac{dy}{dx} = y^{-2/3} \cdot \fracTo find dy/dx, we differentiate the given equation implicitly with respect to x.

Differentiating x = 3y^(1/3) + 2y with respect to x, we get:

1 = (1/3) * 3 * y^(-2/3) * dy/dx + 2 * dy/dx

Simplifying this, we have:

1 = y^(-2/3) * dy/dx + 2 * dy/dx

Multiplying throughout by y^(2/3), we get:

y^(2/3) = dy/dTo find ( \frac{dy}{dx} ) given ( x = 3y^{1/3} + 2y ), differentiate both sides of the equation with respect to ( x ) using implicit differentiation.

Differentiating ( x ) with respect to ( x ) gives ( 1 ). Differentiating ( 3y^{1/3} ) with respect to ( x ) using the chain rule gives ( \frac{1}{3} \cdot 3 \cdot y^{-2/3} \cdot \frac{dy}{dx} = y^{-2/3} \cdot \frac{dyTo find dy/dx, we differentiate the given equation implicitly with respect to x.

Differentiating x = 3y^(1/3) + 2y with respect to x, we get:

1 = (1/3) * 3 * y^(-2/3) * dy/dx + 2 * dy/dx

Simplifying this, we have:

1 = y^(-2/3) * dy/dx + 2 * dy/dx

Multiplying throughout by y^(2/3), we get:

y^(2/3) = dy/dxTo find ( \frac{dy}{dx} ) given ( x = 3y^{1/3} + 2y ), differentiate both sides of the equation with respect to ( x ) using implicit differentiation.

Differentiating ( x ) with respect to ( x ) gives ( 1 ). Differentiating ( 3y^{1/3} ) with respect to ( x ) using the chain rule gives ( \frac{1}{3} \cdot 3 \cdot y^{-2/3} \cdot \frac{dy}{dx} = y^{-2/3} \cdot \frac{dy}{To find dy/dx, we differentiate the given equation implicitly with respect to x.

Differentiating x = 3y^(1/3) + 2y with respect to x, we get:

1 = (1/3) * 3 * y^(-2/3) * dy/dx + 2 * dy/dx

Simplifying this, we have:

1 = y^(-2/3) * dy/dx + 2 * dy/dx

Multiplying throughout by y^(2/3), we get:

y^(2/3) = dy/dx + To find ( \frac{dy}{dx} ) given ( x = 3y^{1/3} + 2y ), differentiate both sides of the equation with respect to ( x ) using implicit differentiation.

Differentiating ( x ) with respect to ( x ) gives ( 1 ). Differentiating ( 3y^{1/3} ) with respect to ( x ) using the chain rule gives ( \frac{1}{3} \cdot 3 \cdot y^{-2/3} \cdot \frac{dy}{dx} = y^{-2/3} \cdot \frac{dy}{dxTo find dy/dx, we differentiate the given equation implicitly with respect to x.

Differentiating x = 3y^(1/3) + 2y with respect to x, we get:

1 = (1/3) * 3 * y^(-2/3) * dy/dx + 2 * dy/dx

Simplifying this, we have:

1 = y^(-2/3) * dy/dx + 2 * dy/dx

Multiplying throughout by y^(2/3), we get:

y^(2/3) = dy/dx + 2To find ( \frac{dy}{dx} ) given ( x = 3y^{1/3} + 2y ), differentiate both sides of the equation with respect to ( x ) using implicit differentiation.

Differentiating ( x ) with respect to ( x ) gives ( 1 ). Differentiating ( 3y^{1/3} ) with respect to ( x ) using the chain rule gives ( \frac{1}{3} \cdot 3 \cdot y^{-2/3} \cdot \frac{dy}{dx} = y^{-2/3} \cdot \frac{dy}{dx} \To find dy/dx, we differentiate the given equation implicitly with respect to x.

Differentiating x = 3y^(1/3) + 2y with respect to x, we get:

1 = (1/3) * 3 * y^(-2/3) * dy/dx + 2 * dy/dx

Simplifying this, we have:

1 = y^(-2/3) * dy/dx + 2 * dy/dx

Multiplying throughout by y^(2/3), we get:

y^(2/3) = dy/dx + 2y^(To find ( \frac{dy}{dx} ) given ( x = 3y^{1/3} + 2y ), differentiate both sides of the equation with respect to ( x ) using implicit differentiation.

Differentiating ( x ) with respect to ( x ) gives ( 1 ). Differentiating ( 3y^{1/3} ) with respect to ( x ) using the chain rule gives ( \frac{1}{3} \cdot 3 \cdot y^{-2/3} \cdot \frac{dy}{dx} = y^{-2/3} \cdot \frac{dy}{dx} ). DifferentTo find dy/dx, we differentiate the given equation implicitly with respect to x.

Differentiating x = 3y^(1/3) + 2y with respect to x, we get:

1 = (1/3) * 3 * y^(-2/3) * dy/dx + 2 * dy/dx

Simplifying this, we have:

1 = y^(-2/3) * dy/dx + 2 * dy/dx

Multiplying throughout by y^(2/3), we get:

y^(2/3) = dy/dx + 2y^(2To find ( \frac{dy}{dx} ) given ( x = 3y^{1/3} + 2y ), differentiate both sides of the equation with respect to ( x ) using implicit differentiation.

Differentiating ( x ) with respect to ( x ) gives ( 1 ). Differentiating ( 3y^{1/3} ) with respect to ( x ) using the chain rule gives ( \frac{1}{3} \cdot 3 \cdot y^{-2/3} \cdot \frac{dy}{dx} = y^{-2/3} \cdot \frac{dy}{dx} ). DifferentiatingTo find dy/dx, we differentiate the given equation implicitly with respect to x.

Differentiating x = 3y^(1/3) + 2y with respect to x, we get:

1 = (1/3) * 3 * y^(-2/3) * dy/dx + 2 * dy/dx

Simplifying this, we have:

1 = y^(-2/3) * dy/dx + 2 * dy/dx

Multiplying throughout by y^(2/3), we get:

y^(2/3) = dy/dx + 2y^(2/To find ( \frac{dy}{dx} ) given ( x = 3y^{1/3} + 2y ), differentiate both sides of the equation with respect to ( x ) using implicit differentiation.

Differentiating ( x ) with respect to ( x ) gives ( 1 ). Differentiating ( 3y^{1/3} ) with respect to ( x ) using the chain rule gives ( \frac{1}{3} \cdot 3 \cdot y^{-2/3} \cdot \frac{dy}{dx} = y^{-2/3} \cdot \frac{dy}{dx} ). Differentiating (To find dy/dx, we differentiate the given equation implicitly with respect to x.

Differentiating x = 3y^(1/3) + 2y with respect to x, we get:

1 = (1/3) * 3 * y^(-2/3) * dy/dx + 2 * dy/dx

Simplifying this, we have:

1 = y^(-2/3) * dy/dx + 2 * dy/dx

Multiplying throughout by y^(2/3), we get:

y^(2/3) = dy/dx + 2y^(2/3)To find ( \frac{dy}{dx} ) given ( x = 3y^{1/3} + 2y ), differentiate both sides of the equation with respect to ( x ) using implicit differentiation.

Differentiating ( x ) with respect to ( x ) gives ( 1 ). Differentiating ( 3y^{1/3} ) with respect to ( x ) using the chain rule gives ( \frac{1}{3} \cdot 3 \cdot y^{-2/3} \cdot \frac{dy}{dx} = y^{-2/3} \cdot \frac{dy}{dx} ). Differentiating ( 2To find dy/dx, we differentiate the given equation implicitly with respect to x.

Differentiating x = 3y^(1/3) + 2y with respect to x, we get:

1 = (1/3) * 3 * y^(-2/3) * dy/dx + 2 * dy/dx

Simplifying this, we have:

1 = y^(-2/3) * dy/dx + 2 * dy/dx

Multiplying throughout by y^(2/3), we get:

y^(2/3) = dy/dx + 2y^(2/3) *To find ( \frac{dy}{dx} ) given ( x = 3y^{1/3} + 2y ), differentiate both sides of the equation with respect to ( x ) using implicit differentiation.

Differentiating ( x ) with respect to ( x ) gives ( 1 ). Differentiating ( 3y^{1/3} ) with respect to ( x ) using the chain rule gives ( \frac{1}{3} \cdot 3 \cdot y^{-2/3} \cdot \frac{dy}{dx} = y^{-2/3} \cdot \frac{dy}{dx} ). Differentiating ( 2yTo find dy/dx, we differentiate the given equation implicitly with respect to x.

Differentiating x = 3y^(1/3) + 2y with respect to x, we get:

1 = (1/3) * 3 * y^(-2/3) * dy/dx + 2 * dy/dx

Simplifying this, we have:

1 = y^(-2/3) * dy/dx + 2 * dy/dx

Multiplying throughout by y^(2/3), we get:

y^(2/3) = dy/dx + 2y^(2/3) * dyTo find ( \frac{dy}{dx} ) given ( x = 3y^{1/3} + 2y ), differentiate both sides of the equation with respect to ( x ) using implicit differentiation.

Differentiating ( x ) with respect to ( x ) gives ( 1 ). Differentiating ( 3y^{1/3} ) with respect to ( x ) using the chain rule gives ( \frac{1}{3} \cdot 3 \cdot y^{-2/3} \cdot \frac{dy}{dx} = y^{-2/3} \cdot \frac{dy}{dx} ). Differentiating ( 2y ) withTo find dy/dx, we differentiate the given equation implicitly with respect to x.

Differentiating x = 3y^(1/3) + 2y with respect to x, we get:

1 = (1/3) * 3 * y^(-2/3) * dy/dx + 2 * dy/dx

Simplifying this, we have:

1 = y^(-2/3) * dy/dx + 2 * dy/dx

Multiplying throughout by y^(2/3), we get:

y^(2/3) = dy/dx + 2y^(2/3) * dy/dxTo find ( \frac{dy}{dx} ) given ( x = 3y^{1/3} + 2y ), differentiate both sides of the equation with respect to ( x ) using implicit differentiation.

Differentiating ( x ) with respect to ( x ) gives ( 1 ). Differentiating ( 3y^{1/3} ) with respect to ( x ) using the chain rule gives ( \frac{1}{3} \cdot 3 \cdot y^{-2/3} \cdot \frac{dy}{dx} = y^{-2/3} \cdot \frac{dy}{dx} ). Differentiating ( 2y ) with respectTo find dy/dx, we differentiate the given equation implicitly with respect to x.

Differentiating x = 3y^(1/3) + 2y with respect to x, we get:

1 = (1/3) * 3 * y^(-2/3) * dy/dx + 2 * dy/dx

Simplifying this, we have:

1 = y^(-2/3) * dy/dx + 2 * dy/dx

Multiplying throughout by y^(2/3), we get:

y^(2/3) = dy/dx + 2y^(2/3) * dy/dx

To find ( \frac{dy}{dx} ) given ( x = 3y^{1/3} + 2y ), differentiate both sides of the equation with respect to ( x ) using implicit differentiation.

Differentiating ( x ) with respect to ( x ) gives ( 1 ). Differentiating ( 3y^{1/3} ) with respect to ( x ) using the chain rule gives ( \frac{1}{3} \cdot 3 \cdot y^{-2/3} \cdot \frac{dy}{dx} = y^{-2/3} \cdot \frac{dy}{dx} ). Differentiating ( 2y ) with respect to (To find dy/dx, we differentiate the given equation implicitly with respect to x.

Differentiating x = 3y^(1/3) + 2y with respect to x, we get:

1 = (1/3) * 3 * y^(-2/3) * dy/dx + 2 * dy/dx

Simplifying this, we have:

1 = y^(-2/3) * dy/dx + 2 * dy/dx

Multiplying throughout by y^(2/3), we get:

y^(2/3) = dy/dx + 2y^(2/3) * dy/dx

Now,To find ( \frac{dy}{dx} ) given ( x = 3y^{1/3} + 2y ), differentiate both sides of the equation with respect to ( x ) using implicit differentiation.

Differentiating ( x ) with respect to ( x ) gives ( 1 ). Differentiating ( 3y^{1/3} ) with respect to ( x ) using the chain rule gives ( \frac{1}{3} \cdot 3 \cdot y^{-2/3} \cdot \frac{dy}{dx} = y^{-2/3} \cdot \frac{dy}{dx} ). Differentiating ( 2y ) with respect to ( xTo find dy/dx, we differentiate the given equation implicitly with respect to x.

Differentiating x = 3y^(1/3) + 2y with respect to x, we get:

1 = (1/3) * 3 * y^(-2/3) * dy/dx + 2 * dy/dx

Simplifying this, we have:

1 = y^(-2/3) * dy/dx + 2 * dy/dx

Multiplying throughout by y^(2/3), we get:

y^(2/3) = dy/dx + 2y^(2/3) * dy/dx

Now, combiningTo find ( \frac{dy}{dx} ) given ( x = 3y^{1/3} + 2y ), differentiate both sides of the equation with respect to ( x ) using implicit differentiation.

Differentiating ( x ) with respect to ( x ) gives ( 1 ). Differentiating ( 3y^{1/3} ) with respect to ( x ) using the chain rule gives ( \frac{1}{3} \cdot 3 \cdot y^{-2/3} \cdot \frac{dy}{dx} = y^{-2/3} \cdot \frac{dy}{dx} ). Differentiating ( 2y ) with respect to ( x \To find dy/dx, we differentiate the given equation implicitly with respect to x.

Differentiating x = 3y^(1/3) + 2y with respect to x, we get:

1 = (1/3) * 3 * y^(-2/3) * dy/dx + 2 * dy/dx

Simplifying this, we have:

1 = y^(-2/3) * dy/dx + 2 * dy/dx

Multiplying throughout by y^(2/3), we get:

y^(2/3) = dy/dx + 2y^(2/3) * dy/dx

Now, combining likeTo find ( \frac{dy}{dx} ) given ( x = 3y^{1/3} + 2y ), differentiate both sides of the equation with respect to ( x ) using implicit differentiation.

Differentiating ( x ) with respect to ( x ) gives ( 1 ). Differentiating ( 3y^{1/3} ) with respect to ( x ) using the chain rule gives ( \frac{1}{3} \cdot 3 \cdot y^{-2/3} \cdot \frac{dy}{dx} = y^{-2/3} \cdot \frac{dy}{dx} ). Differentiating ( 2y ) with respect to ( x )To find dy/dx, we differentiate the given equation implicitly with respect to x.

Differentiating x = 3y^(1/3) + 2y with respect to x, we get:

1 = (1/3) * 3 * y^(-2/3) * dy/dx + 2 * dy/dx

Simplifying this, we have:

1 = y^(-2/3) * dy/dx + 2 * dy/dx

Multiplying throughout by y^(2/3), we get:

y^(2/3) = dy/dx + 2y^(2/3) * dy/dx

Now, combining like termsTo find ( \frac{dy}{dx} ) given ( x = 3y^{1/3} + 2y ), differentiate both sides of the equation with respect to ( x ) using implicit differentiation.

Differentiating ( x ) with respect to ( x ) gives ( 1 ). Differentiating ( 3y^{1/3} ) with respect to ( x ) using the chain rule gives ( \frac{1}{3} \cdot 3 \cdot y^{-2/3} \cdot \frac{dy}{dx} = y^{-2/3} \cdot \frac{dy}{dx} ). Differentiating ( 2y ) with respect to ( x ) givesTo find dy/dx, we differentiate the given equation implicitly with respect to x.

Differentiating x = 3y^(1/3) + 2y with respect to x, we get:

1 = (1/3) * 3 * y^(-2/3) * dy/dx + 2 * dy/dx

Simplifying this, we have:

1 = y^(-2/3) * dy/dx + 2 * dy/dx

Multiplying throughout by y^(2/3), we get:

y^(2/3) = dy/dx + 2y^(2/3) * dy/dx

Now, combining like terms,To find ( \frac{dy}{dx} ) given ( x = 3y^{1/3} + 2y ), differentiate both sides of the equation with respect to ( x ) using implicit differentiation.

Differentiating ( x ) with respect to ( x ) gives ( 1 ). Differentiating ( 3y^{1/3} ) with respect to ( x ) using the chain rule gives ( \frac{1}{3} \cdot 3 \cdot y^{-2/3} \cdot \frac{dy}{dx} = y^{-2/3} \cdot \frac{dy}{dx} ). Differentiating ( 2y ) with respect to ( x ) gives (To find dy/dx, we differentiate the given equation implicitly with respect to x.

Differentiating x = 3y^(1/3) + 2y with respect to x, we get:

1 = (1/3) * 3 * y^(-2/3) * dy/dx + 2 * dy/dx

Simplifying this, we have:

1 = y^(-2/3) * dy/dx + 2 * dy/dx

Multiplying throughout by y^(2/3), we get:

y^(2/3) = dy/dx + 2y^(2/3) * dy/dx

Now, combining like terms, weTo find ( \frac{dy}{dx} ) given ( x = 3y^{1/3} + 2y ), differentiate both sides of the equation with respect to ( x ) using implicit differentiation.

Differentiating ( x ) with respect to ( x ) gives ( 1 ). Differentiating ( 3y^{1/3} ) with respect to ( x ) using the chain rule gives ( \frac{1}{3} \cdot 3 \cdot y^{-2/3} \cdot \frac{dy}{dx} = y^{-2/3} \cdot \frac{dy}{dx} ). Differentiating ( 2y ) with respect to ( x ) gives ( 2To find dy/dx, we differentiate the given equation implicitly with respect to x.

Differentiating x = 3y^(1/3) + 2y with respect to x, we get:

1 = (1/3) * 3 * y^(-2/3) * dy/dx + 2 * dy/dx

Simplifying this, we have:

1 = y^(-2/3) * dy/dx + 2 * dy/dx

Multiplying throughout by y^(2/3), we get:

y^(2/3) = dy/dx + 2y^(2/3) * dy/dx

Now, combining like terms, we get:

To find ( \frac{dy}{dx} ) given ( x = 3y^{1/3} + 2y ), differentiate both sides of the equation with respect to ( x ) using implicit differentiation.

Differentiating ( x ) with respect to ( x ) gives ( 1 ). Differentiating ( 3y^{1/3} ) with respect to ( x ) using the chain rule gives ( \frac{1}{3} \cdot 3 \cdot y^{-2/3} \cdot \frac{dy}{dx} = y^{-2/3} \cdot \frac{dy}{dx} ). Differentiating ( 2y ) with respect to ( x ) gives ( 2 \fracTo find dy/dx, we differentiate the given equation implicitly with respect to x.

Differentiating x = 3y^(1/3) + 2y with respect to x, we get:

1 = (1/3) * 3 * y^(-2/3) * dy/dx + 2 * dy/dx

Simplifying this, we have:

1 = y^(-2/3) * dy/dx + 2 * dy/dx

Multiplying throughout by y^(2/3), we get:

y^(2/3) = dy/dx + 2y^(2/3) * dy/dx

Now, combining like terms, we get:

y^(To find ( \frac{dy}{dx} ) given ( x = 3y^{1/3} + 2y ), differentiate both sides of the equation with respect to ( x ) using implicit differentiation.

Differentiating ( x ) with respect to ( x ) gives ( 1 ). Differentiating ( 3y^{1/3} ) with respect to ( x ) using the chain rule gives ( \frac{1}{3} \cdot 3 \cdot y^{-2/3} \cdot \frac{dy}{dx} = y^{-2/3} \cdot \frac{dy}{dx} ). Differentiating ( 2y ) with respect to ( x ) gives ( 2 \frac{dyTo find dy/dx, we differentiate the given equation implicitly with respect to x.

Differentiating x = 3y^(1/3) + 2y with respect to x, we get:

1 = (1/3) * 3 * y^(-2/3) * dy/dx + 2 * dy/dx

Simplifying this, we have:

1 = y^(-2/3) * dy/dx + 2 * dy/dx

Multiplying throughout by y^(2/3), we get:

y^(2/3) = dy/dx + 2y^(2/3) * dy/dx

Now, combining like terms, we get:

y^(2/To find ( \frac{dy}{dx} ) given ( x = 3y^{1/3} + 2y ), differentiate both sides of the equation with respect to ( x ) using implicit differentiation.

Differentiating ( x ) with respect to ( x ) gives ( 1 ). Differentiating ( 3y^{1/3} ) with respect to ( x ) using the chain rule gives ( \frac{1}{3} \cdot 3 \cdot y^{-2/3} \cdot \frac{dy}{dx} = y^{-2/3} \cdot \frac{dy}{dx} ). Differentiating ( 2y ) with respect to ( x ) gives ( 2 \frac{dy}{dxTo find dy/dx, we differentiate the given equation implicitly with respect to x.

Differentiating x = 3y^(1/3) + 2y with respect to x, we get:

1 = (1/3) * 3 * y^(-2/3) * dy/dx + 2 * dy/dx

Simplifying this, we have:

1 = y^(-2/3) * dy/dx + 2 * dy/dx

Multiplying throughout by y^(2/3), we get:

y^(2/3) = dy/dx + 2y^(2/3) * dy/dx

Now, combining like terms, we get:

y^(2/3)To find ( \frac{dy}{dx} ) given ( x = 3y^{1/3} + 2y ), differentiate both sides of the equation with respect to ( x ) using implicit differentiation.

Differentiating ( x ) with respect to ( x ) gives ( 1 ). Differentiating ( 3y^{1/3} ) with respect to ( x ) using the chain rule gives ( \frac{1}{3} \cdot 3 \cdot y^{-2/3} \cdot \frac{dy}{dx} = y^{-2/3} \cdot \frac{dy}{dx} ). Differentiating ( 2y ) with respect to ( x ) gives ( 2 \frac{dy}{dx} \To find dy/dx, we differentiate the given equation implicitly with respect to x.

Differentiating x = 3y^(1/3) + 2y with respect to x, we get:

1 = (1/3) * 3 * y^(-2/3) * dy/dx + 2 * dy/dx

Simplifying this, we have:

1 = y^(-2/3) * dy/dx + 2 * dy/dx

Multiplying throughout by y^(2/3), we get:

y^(2/3) = dy/dx + 2y^(2/3) * dy/dx

Now, combining like terms, we get:

y^(2/3) =To find ( \frac{dy}{dx} ) given ( x = 3y^{1/3} + 2y ), differentiate both sides of the equation with respect to ( x ) using implicit differentiation.

Differentiating ( x ) with respect to ( x ) gives ( 1 ). Differentiating ( 3y^{1/3} ) with respect to ( x ) using the chain rule gives ( \frac{1}{3} \cdot 3 \cdot y^{-2/3} \cdot \frac{dy}{dx} = y^{-2/3} \cdot \frac{dy}{dx} ). Differentiating ( 2y ) with respect to ( x ) gives ( 2 \frac{dy}{dx} ).

PuttingTo find dy/dx, we differentiate the given equation implicitly with respect to x.

Differentiating x = 3y^(1/3) + 2y with respect to x, we get:

1 = (1/3) * 3 * y^(-2/3) * dy/dx + 2 * dy/dx

Simplifying this, we have:

1 = y^(-2/3) * dy/dx + 2 * dy/dx

Multiplying throughout by y^(2/3), we get:

y^(2/3) = dy/dx + 2y^(2/3) * dy/dx

Now, combining like terms, we get:

y^(2/3) = (To find ( \frac{dy}{dx} ) given ( x = 3y^{1/3} + 2y ), differentiate both sides of the equation with respect to ( x ) using implicit differentiation.

Differentiating ( x ) with respect to ( x ) gives ( 1 ). Differentiating ( 3y^{1/3} ) with respect to ( x ) using the chain rule gives ( \frac{1}{3} \cdot 3 \cdot y^{-2/3} \cdot \frac{dy}{dx} = y^{-2/3} \cdot \frac{dy}{dx} ). Differentiating ( 2y ) with respect to ( x ) gives ( 2 \frac{dy}{dx} ).

Putting these togetherTo find dy/dx, we differentiate the given equation implicitly with respect to x.

Differentiating x = 3y^(1/3) + 2y with respect to x, we get:

1 = (1/3) * 3 * y^(-2/3) * dy/dx + 2 * dy/dx

Simplifying this, we have:

1 = y^(-2/3) * dy/dx + 2 * dy/dx

Multiplying throughout by y^(2/3), we get:

y^(2/3) = dy/dx + 2y^(2/3) * dy/dx

Now, combining like terms, we get:

y^(2/3) = (1 +To find ( \frac{dy}{dx} ) given ( x = 3y^{1/3} + 2y ), differentiate both sides of the equation with respect to ( x ) using implicit differentiation.

Differentiating ( x ) with respect to ( x ) gives ( 1 ). Differentiating ( 3y^{1/3} ) with respect to ( x ) using the chain rule gives ( \frac{1}{3} \cdot 3 \cdot y^{-2/3} \cdot \frac{dy}{dx} = y^{-2/3} \cdot \frac{dy}{dx} ). Differentiating ( 2y ) with respect to ( x ) gives ( 2 \frac{dy}{dx} ).

Putting these together,To find dy/dx, we differentiate the given equation implicitly with respect to x.

Differentiating x = 3y^(1/3) + 2y with respect to x, we get:

1 = (1/3) * 3 * y^(-2/3) * dy/dx + 2 * dy/dx

Simplifying this, we have:

1 = y^(-2/3) * dy/dx + 2 * dy/dx

Multiplying throughout by y^(2/3), we get:

y^(2/3) = dy/dx + 2y^(2/3) * dy/dx

Now, combining like terms, we get:

y^(2/3) = (1 + To find ( \frac{dy}{dx} ) given ( x = 3y^{1/3} + 2y ), differentiate both sides of the equation with respect to ( x ) using implicit differentiation.

Differentiating ( x ) with respect to ( x ) gives ( 1 ). Differentiating ( 3y^{1/3} ) with respect to ( x ) using the chain rule gives ( \frac{1}{3} \cdot 3 \cdot y^{-2/3} \cdot \frac{dy}{dx} = y^{-2/3} \cdot \frac{dy}{dx} ). Differentiating ( 2y ) with respect to ( x ) gives ( 2 \frac{dy}{dx} ).

Putting these together, we have (To find dy/dx, we differentiate the given equation implicitly with respect to x.

Differentiating x = 3y^(1/3) + 2y with respect to x, we get:

1 = (1/3) * 3 * y^(-2/3) * dy/dx + 2 * dy/dx

Simplifying this, we have:

1 = y^(-2/3) * dy/dx + 2 * dy/dx

Multiplying throughout by y^(2/3), we get:

y^(2/3) = dy/dx + 2y^(2/3) * dy/dx

Now, combining like terms, we get:

y^(2/3) = (1 + 2y^(To find ( \frac{dy}{dx} ) given ( x = 3y^{1/3} + 2y ), differentiate both sides of the equation with respect to ( x ) using implicit differentiation.

Differentiating ( x ) with respect to ( x ) gives ( 1 ). Differentiating ( 3y^{1/3} ) with respect to ( x ) using the chain rule gives ( \frac{1}{3} \cdot 3 \cdot y^{-2/3} \cdot \frac{dy}{dx} = y^{-2/3} \cdot \frac{dy}{dx} ). Differentiating ( 2y ) with respect to ( x ) gives ( 2 \frac{dy}{dx} ).

Putting these together, we have ( To find dy/dx, we differentiate the given equation implicitly with respect to x.

Differentiating x = 3y^(1/3) + 2y with respect to x, we get:

1 = (1/3) * 3 * y^(-2/3) * dy/dx + 2 * dy/dx

Simplifying this, we have:

1 = y^(-2/3) * dy/dx + 2 * dy/dx

Multiplying throughout by y^(2/3), we get:

y^(2/3) = dy/dx + 2y^(2/3) * dy/dx

Now, combining like terms, we get:

y^(2/3) = (1 + 2y^(2To find ( \frac{dy}{dx} ) given ( x = 3y^{1/3} + 2y ), differentiate both sides of the equation with respect to ( x ) using implicit differentiation.

Differentiating ( x ) with respect to ( x ) gives ( 1 ). Differentiating ( 3y^{1/3} ) with respect to ( x ) using the chain rule gives ( \frac{1}{3} \cdot 3 \cdot y^{-2/3} \cdot \frac{dy}{dx} = y^{-2/3} \cdot \frac{dy}{dx} ). Differentiating ( 2y ) with respect to ( x ) gives ( 2 \frac{dy}{dx} ).

Putting these together, we have ( 1To find dy/dx, we differentiate the given equation implicitly with respect to x.

Differentiating x = 3y^(1/3) + 2y with respect to x, we get:

1 = (1/3) * 3 * y^(-2/3) * dy/dx + 2 * dy/dx

Simplifying this, we have:

1 = y^(-2/3) * dy/dx + 2 * dy/dx

Multiplying throughout by y^(2/3), we get:

y^(2/3) = dy/dx + 2y^(2/3) * dy/dx

Now, combining like terms, we get:

y^(2/3) = (1 + 2y^(2/To find ( \frac{dy}{dx} ) given ( x = 3y^{1/3} + 2y ), differentiate both sides of the equation with respect to ( x ) using implicit differentiation.

Differentiating ( x ) with respect to ( x ) gives ( 1 ). Differentiating ( 3y^{1/3} ) with respect to ( x ) using the chain rule gives ( \frac{1}{3} \cdot 3 \cdot y^{-2/3} \cdot \frac{dy}{dx} = y^{-2/3} \cdot \frac{dy}{dx} ). Differentiating ( 2y ) with respect to ( x ) gives ( 2 \frac{dy}{dx} ).

Putting these together, we have ( 1 =To find dy/dx, we differentiate the given equation implicitly with respect to x.

Differentiating x = 3y^(1/3) + 2y with respect to x, we get:

1 = (1/3) * 3 * y^(-2/3) * dy/dx + 2 * dy/dx

Simplifying this, we have:

1 = y^(-2/3) * dy/dx + 2 * dy/dx

Multiplying throughout by y^(2/3), we get:

y^(2/3) = dy/dx + 2y^(2/3) * dy/dx

Now, combining like terms, we get:

y^(2/3) = (1 + 2y^(2/3To find ( \frac{dy}{dx} ) given ( x = 3y^{1/3} + 2y ), differentiate both sides of the equation with respect to ( x ) using implicit differentiation.

Differentiating ( x ) with respect to ( x ) gives ( 1 ). Differentiating ( 3y^{1/3} ) with respect to ( x ) using the chain rule gives ( \frac{1}{3} \cdot 3 \cdot y^{-2/3} \cdot \frac{dy}{dx} = y^{-2/3} \cdot \frac{dy}{dx} ). Differentiating ( 2y ) with respect to ( x ) gives ( 2 \frac{dy}{dx} ).

Putting these together, we have ( 1 = yTo find dy/dx, we differentiate the given equation implicitly with respect to x.

Differentiating x = 3y^(1/3) + 2y with respect to x, we get:

1 = (1/3) * 3 * y^(-2/3) * dy/dx + 2 * dy/dx

Simplifying this, we have:

1 = y^(-2/3) * dy/dx + 2 * dy/dx

Multiplying throughout by y^(2/3), we get:

y^(2/3) = dy/dx + 2y^(2/3) * dy/dx

Now, combining like terms, we get:

y^(2/3) = (1 + 2y^(2/3))To find ( \frac{dy}{dx} ) given ( x = 3y^{1/3} + 2y ), differentiate both sides of the equation with respect to ( x ) using implicit differentiation.

Differentiating ( x ) with respect to ( x ) gives ( 1 ). Differentiating ( 3y^{1/3} ) with respect to ( x ) using the chain rule gives ( \frac{1}{3} \cdot 3 \cdot y^{-2/3} \cdot \frac{dy}{dx} = y^{-2/3} \cdot \frac{dy}{dx} ). Differentiating ( 2y ) with respect to ( x ) gives ( 2 \frac{dy}{dx} ).

Putting these together, we have ( 1 = y^{-To find dy/dx, we differentiate the given equation implicitly with respect to x.

Differentiating x = 3y^(1/3) + 2y with respect to x, we get:

1 = (1/3) * 3 * y^(-2/3) * dy/dx + 2 * dy/dx

Simplifying this, we have:

1 = y^(-2/3) * dy/dx + 2 * dy/dx

Multiplying throughout by y^(2/3), we get:

y^(2/3) = dy/dx + 2y^(2/3) * dy/dx

Now, combining like terms, we get:

y^(2/3) = (1 + 2y^(2/3)) *To find ( \frac{dy}{dx} ) given ( x = 3y^{1/3} + 2y ), differentiate both sides of the equation with respect to ( x ) using implicit differentiation.

Differentiating ( x ) with respect to ( x ) gives ( 1 ). Differentiating ( 3y^{1/3} ) with respect to ( x ) using the chain rule gives ( \frac{1}{3} \cdot 3 \cdot y^{-2/3} \cdot \frac{dy}{dx} = y^{-2/3} \cdot \frac{dy}{dx} ). Differentiating ( 2y ) with respect to ( x ) gives ( 2 \frac{dy}{dx} ).

Putting these together, we have ( 1 = y^{-2To find dy/dx, we differentiate the given equation implicitly with respect to x.

Differentiating x = 3y^(1/3) + 2y with respect to x, we get:

1 = (1/3) * 3 * y^(-2/3) * dy/dx + 2 * dy/dx

Simplifying this, we have:

1 = y^(-2/3) * dy/dx + 2 * dy/dx

Multiplying throughout by y^(2/3), we get:

y^(2/3) = dy/dx + 2y^(2/3) * dy/dx

Now, combining like terms, we get:

y^(2/3) = (1 + 2y^(2/3)) * dyTo find ( \frac{dy}{dx} ) given ( x = 3y^{1/3} + 2y ), differentiate both sides of the equation with respect to ( x ) using implicit differentiation.

Differentiating ( x ) with respect to ( x ) gives ( 1 ). Differentiating ( 3y^{1/3} ) with respect to ( x ) using the chain rule gives ( \frac{1}{3} \cdot 3 \cdot y^{-2/3} \cdot \frac{dy}{dx} = y^{-2/3} \cdot \frac{dy}{dx} ). Differentiating ( 2y ) with respect to ( x ) gives ( 2 \frac{dy}{dx} ).

Putting these together, we have ( 1 = y^{-2/To find dy/dx, we differentiate the given equation implicitly with respect to x.

Differentiating x = 3y^(1/3) + 2y with respect to x, we get:

1 = (1/3) * 3 * y^(-2/3) * dy/dx + 2 * dy/dx

Simplifying this, we have:

1 = y^(-2/3) * dy/dx + 2 * dy/dx

Multiplying throughout by y^(2/3), we get:

y^(2/3) = dy/dx + 2y^(2/3) * dy/dx

Now, combining like terms, we get:

y^(2/3) = (1 + 2y^(2/3)) * dy/dTo find ( \frac{dy}{dx} ) given ( x = 3y^{1/3} + 2y ), differentiate both sides of the equation with respect to ( x ) using implicit differentiation.

Differentiating ( x ) with respect to ( x ) gives ( 1 ). Differentiating ( 3y^{1/3} ) with respect to ( x ) using the chain rule gives ( \frac{1}{3} \cdot 3 \cdot y^{-2/3} \cdot \frac{dy}{dx} = y^{-2/3} \cdot \frac{dy}{dx} ). Differentiating ( 2y ) with respect to ( x ) gives ( 2 \frac{dy}{dx} ).

Putting these together, we have ( 1 = y^{-2/3To find dy/dx, we differentiate the given equation implicitly with respect to x.

Differentiating x = 3y^(1/3) + 2y with respect to x, we get:

1 = (1/3) * 3 * y^(-2/3) * dy/dx + 2 * dy/dx

Simplifying this, we have:

1 = y^(-2/3) * dy/dx + 2 * dy/dx

Multiplying throughout by y^(2/3), we get:

y^(2/3) = dy/dx + 2y^(2/3) * dy/dx

Now, combining like terms, we get:

y^(2/3) = (1 + 2y^(2/3)) * dy/dxTo find ( \frac{dy}{dx} ) given ( x = 3y^{1/3} + 2y ), differentiate both sides of the equation with respect to ( x ) using implicit differentiation.

Differentiating ( x ) with respect to ( x ) gives ( 1 ). Differentiating ( 3y^{1/3} ) with respect to ( x ) using the chain rule gives ( \frac{1}{3} \cdot 3 \cdot y^{-2/3} \cdot \frac{dy}{dx} = y^{-2/3} \cdot \frac{dy}{dx} ). Differentiating ( 2y ) with respect to ( x ) gives ( 2 \frac{dy}{dx} ).

Putting these together, we have ( 1 = y^{-2/3}To find dy/dx, we differentiate the given equation implicitly with respect to x.

Differentiating x = 3y^(1/3) + 2y with respect to x, we get:

1 = (1/3) * 3 * y^(-2/3) * dy/dx + 2 * dy/dx

Simplifying this, we have:

1 = y^(-2/3) * dy/dx + 2 * dy/dx

Multiplying throughout by y^(2/3), we get:

y^(2/3) = dy/dx + 2y^(2/3) * dy/dx

Now, combining like terms, we get:

y^(2/3) = (1 + 2y^(2/3)) * dy/dx

To find ( \frac{dy}{dx} ) given ( x = 3y^{1/3} + 2y ), differentiate both sides of the equation with respect to ( x ) using implicit differentiation.

Differentiating ( x ) with respect to ( x ) gives ( 1 ). Differentiating ( 3y^{1/3} ) with respect to ( x ) using the chain rule gives ( \frac{1}{3} \cdot 3 \cdot y^{-2/3} \cdot \frac{dy}{dx} = y^{-2/3} \cdot \frac{dy}{dx} ). Differentiating ( 2y ) with respect to ( x ) gives ( 2 \frac{dy}{dx} ).

Putting these together, we have ( 1 = y^{-2/3} \To find dy/dx, we differentiate the given equation implicitly with respect to x.

Differentiating x = 3y^(1/3) + 2y with respect to x, we get:

1 = (1/3) * 3 * y^(-2/3) * dy/dx + 2 * dy/dx

Simplifying this, we have:

1 = y^(-2/3) * dy/dx + 2 * dy/dx

Multiplying throughout by y^(2/3), we get:

y^(2/3) = dy/dx + 2y^(2/3) * dy/dx

Now, combining like terms, we get:

y^(2/3) = (1 + 2y^(2/3)) * dy/dx

SoTo find ( \frac{dy}{dx} ) given ( x = 3y^{1/3} + 2y ), differentiate both sides of the equation with respect to ( x ) using implicit differentiation.

Differentiating ( x ) with respect to ( x ) gives ( 1 ). Differentiating ( 3y^{1/3} ) with respect to ( x ) using the chain rule gives ( \frac{1}{3} \cdot 3 \cdot y^{-2/3} \cdot \frac{dy}{dx} = y^{-2/3} \cdot \frac{dy}{dx} ). Differentiating ( 2y ) with respect to ( x ) gives ( 2 \frac{dy}{dx} ).

Putting these together, we have ( 1 = y^{-2/3} \cdotTo find dy/dx, we differentiate the given equation implicitly with respect to x.

Differentiating x = 3y^(1/3) + 2y with respect to x, we get:

1 = (1/3) * 3 * y^(-2/3) * dy/dx + 2 * dy/dx

Simplifying this, we have:

1 = y^(-2/3) * dy/dx + 2 * dy/dx

Multiplying throughout by y^(2/3), we get:

y^(2/3) = dy/dx + 2y^(2/3) * dy/dx

Now, combining like terms, we get:

y^(2/3) = (1 + 2y^(2/3)) * dy/dx

So,To find ( \frac{dy}{dx} ) given ( x = 3y^{1/3} + 2y ), differentiate both sides of the equation with respect to ( x ) using implicit differentiation.

Differentiating ( x ) with respect to ( x ) gives ( 1 ). Differentiating ( 3y^{1/3} ) with respect to ( x ) using the chain rule gives ( \frac{1}{3} \cdot 3 \cdot y^{-2/3} \cdot \frac{dy}{dx} = y^{-2/3} \cdot \frac{dy}{dx} ). Differentiating ( 2y ) with respect to ( x ) gives ( 2 \frac{dy}{dx} ).

Putting these together, we have ( 1 = y^{-2/3} \cdot \To find dy/dx, we differentiate the given equation implicitly with respect to x.

Differentiating x = 3y^(1/3) + 2y with respect to x, we get:

1 = (1/3) * 3 * y^(-2/3) * dy/dx + 2 * dy/dx

Simplifying this, we have:

1 = y^(-2/3) * dy/dx + 2 * dy/dx

Multiplying throughout by y^(2/3), we get:

y^(2/3) = dy/dx + 2y^(2/3) * dy/dx

Now, combining like terms, we get:

y^(2/3) = (1 + 2y^(2/3)) * dy/dx

So, finallyTo find ( \frac{dy}{dx} ) given ( x = 3y^{1/3} + 2y ), differentiate both sides of the equation with respect to ( x ) using implicit differentiation.

Differentiating ( x ) with respect to ( x ) gives ( 1 ). Differentiating ( 3y^{1/3} ) with respect to ( x ) using the chain rule gives ( \frac{1}{3} \cdot 3 \cdot y^{-2/3} \cdot \frac{dy}{dx} = y^{-2/3} \cdot \frac{dy}{dx} ). Differentiating ( 2y ) with respect to ( x ) gives ( 2 \frac{dy}{dx} ).

Putting these together, we have ( 1 = y^{-2/3} \cdot \fracTo find dy/dx, we differentiate the given equation implicitly with respect to x.

Differentiating x = 3y^(1/3) + 2y with respect to x, we get:

1 = (1/3) * 3 * y^(-2/3) * dy/dx + 2 * dy/dx

Simplifying this, we have:

1 = y^(-2/3) * dy/dx + 2 * dy/dx

Multiplying throughout by y^(2/3), we get:

y^(2/3) = dy/dx + 2y^(2/3) * dy/dx

Now, combining like terms, we get:

y^(2/3) = (1 + 2y^(2/3)) * dy/dx

So, finally,To find ( \frac{dy}{dx} ) given ( x = 3y^{1/3} + 2y ), differentiate both sides of the equation with respect to ( x ) using implicit differentiation.

Differentiating ( x ) with respect to ( x ) gives ( 1 ). Differentiating ( 3y^{1/3} ) with respect to ( x ) using the chain rule gives ( \frac{1}{3} \cdot 3 \cdot y^{-2/3} \cdot \frac{dy}{dx} = y^{-2/3} \cdot \frac{dy}{dx} ). Differentiating ( 2y ) with respect to ( x ) gives ( 2 \frac{dy}{dx} ).

Putting these together, we have ( 1 = y^{-2/3} \cdot \frac{To find dy/dx, we differentiate the given equation implicitly with respect to x.

Differentiating x = 3y^(1/3) + 2y with respect to x, we get:

1 = (1/3) * 3 * y^(-2/3) * dy/dx + 2 * dy/dx

Simplifying this, we have:

1 = y^(-2/3) * dy/dx + 2 * dy/dx

Multiplying throughout by y^(2/3), we get:

y^(2/3) = dy/dx + 2y^(2/3) * dy/dx

Now, combining like terms, we get:

y^(2/3) = (1 + 2y^(2/3)) * dy/dx

So, finally, weTo find ( \frac{dy}{dx} ) given ( x = 3y^{1/3} + 2y ), differentiate both sides of the equation with respect to ( x ) using implicit differentiation.

Differentiating ( x ) with respect to ( x ) gives ( 1 ). Differentiating ( 3y^{1/3} ) with respect to ( x ) using the chain rule gives ( \frac{1}{3} \cdot 3 \cdot y^{-2/3} \cdot \frac{dy}{dx} = y^{-2/3} \cdot \frac{dy}{dx} ). Differentiating ( 2y ) with respect to ( x ) gives ( 2 \frac{dy}{dx} ).

Putting these together, we have ( 1 = y^{-2/3} \cdot \frac{dy}{dxTo find dy/dx, we differentiate the given equation implicitly with respect to x.

Differentiating x = 3y^(1/3) + 2y with respect to x, we get:

1 = (1/3) * 3 * y^(-2/3) * dy/dx + 2 * dy/dx

Simplifying this, we have:

1 = y^(-2/3) * dy/dx + 2 * dy/dx

Multiplying throughout by y^(2/3), we get:

y^(2/3) = dy/dx + 2y^(2/3) * dy/dx

Now, combining like terms, we get:

y^(2/3) = (1 + 2y^(2/3)) * dy/dx

So, finally, we can solve forTo find ( \frac{dy}{dx} ) given ( x = 3y^{1/3} + 2y ), differentiate both sides of the equation with respect to ( x ) using implicit differentiation.

Differentiating ( x ) with respect to ( x ) gives ( 1 ). Differentiating ( 3y^{1/3} ) with respect to ( x ) using the chain rule gives ( \frac{1}{3} \cdot 3 \cdot y^{-2/3} \cdot \frac{dy}{dx} = y^{-2/3} \cdot \frac{dy}{dx} ). Differentiating ( 2y ) with respect to ( x ) gives ( 2 \frac{dy}{dx} ).

Putting these together, we have ( 1 = y^{-2/3} \cdot \frac{dy}{dx}To find dy/dx, we differentiate the given equation implicitly with respect to x.

Differentiating x = 3y^(1/3) + 2y with respect to x, we get:

1 = (1/3) * 3 * y^(-2/3) * dy/dx + 2 * dy/dx

Simplifying this, we have:

1 = y^(-2/3) * dy/dx + 2 * dy/dx

Multiplying throughout by y^(2/3), we get:

y^(2/3) = dy/dx + 2y^(2/3) * dy/dx

Now, combining like terms, we get:

y^(2/3) = (1 + 2y^(2/3)) * dy/dx

So, finally, we can solve for dyTo find ( \frac{dy}{dx} ) given ( x = 3y^{1/3} + 2y ), differentiate both sides of the equation with respect to ( x ) using implicit differentiation.

Differentiating ( x ) with respect to ( x ) gives ( 1 ). Differentiating ( 3y^{1/3} ) with respect to ( x ) using the chain rule gives ( \frac{1}{3} \cdot 3 \cdot y^{-2/3} \cdot \frac{dy}{dx} = y^{-2/3} \cdot \frac{dy}{dx} ). Differentiating ( 2y ) with respect to ( x ) gives ( 2 \frac{dy}{dx} ).

Putting these together, we have ( 1 = y^{-2/3} \cdot \frac{dy}{dx} +To find dy/dx, we differentiate the given equation implicitly with respect to x.

Differentiating x = 3y^(1/3) + 2y with respect to x, we get:

1 = (1/3) * 3 * y^(-2/3) * dy/dx + 2 * dy/dx

Simplifying this, we have:

1 = y^(-2/3) * dy/dx + 2 * dy/dx

Multiplying throughout by y^(2/3), we get:

y^(2/3) = dy/dx + 2y^(2/3) * dy/dx

Now, combining like terms, we get:

y^(2/3) = (1 + 2y^(2/3)) * dy/dx

So, finally, we can solve for dy/dTo find ( \frac{dy}{dx} ) given ( x = 3y^{1/3} + 2y ), differentiate both sides of the equation with respect to ( x ) using implicit differentiation.

Differentiating ( x ) with respect to ( x ) gives ( 1 ). Differentiating ( 3y^{1/3} ) with respect to ( x ) using the chain rule gives ( \frac{1}{3} \cdot 3 \cdot y^{-2/3} \cdot \frac{dy}{dx} = y^{-2/3} \cdot \frac{dy}{dx} ). Differentiating ( 2y ) with respect to ( x ) gives ( 2 \frac{dy}{dx} ).

Putting these together, we have ( 1 = y^{-2/3} \cdot \frac{dy}{dx} + To find dy/dx, we differentiate the given equation implicitly with respect to x.

Differentiating x = 3y^(1/3) + 2y with respect to x, we get:

1 = (1/3) * 3 * y^(-2/3) * dy/dx + 2 * dy/dx

Simplifying this, we have:

1 = y^(-2/3) * dy/dx + 2 * dy/dx

Multiplying throughout by y^(2/3), we get:

y^(2/3) = dy/dx + 2y^(2/3) * dy/dx

Now, combining like terms, we get:

y^(2/3) = (1 + 2y^(2/3)) * dy/dx

So, finally, we can solve for dy/dxTo find ( \frac{dy}{dx} ) given ( x = 3y^{1/3} + 2y ), differentiate both sides of the equation with respect to ( x ) using implicit differentiation.

Differentiating ( x ) with respect to ( x ) gives ( 1 ). Differentiating ( 3y^{1/3} ) with respect to ( x ) using the chain rule gives ( \frac{1}{3} \cdot 3 \cdot y^{-2/3} \cdot \frac{dy}{dx} = y^{-2/3} \cdot \frac{dy}{dx} ). Differentiating ( 2y ) with respect to ( x ) gives ( 2 \frac{dy}{dx} ).

Putting these together, we have ( 1 = y^{-2/3} \cdot \frac{dy}{dx} + 2To find dy/dx, we differentiate the given equation implicitly with respect to x.

Differentiating x = 3y^(1/3) + 2y with respect to x, we get:

1 = (1/3) * 3 * y^(-2/3) * dy/dx + 2 * dy/dx

Simplifying this, we have:

1 = y^(-2/3) * dy/dx + 2 * dy/dx

Multiplying throughout by y^(2/3), we get:

y^(2/3) = dy/dx + 2y^(2/3) * dy/dx

Now, combining like terms, we get:

y^(2/3) = (1 + 2y^(2/3)) * dy/dx

So, finally, we can solve for dy/dx:

To find ( \frac{dy}{dx} ) given ( x = 3y^{1/3} + 2y ), differentiate both sides of the equation with respect to ( x ) using implicit differentiation.

Differentiating ( x ) with respect to ( x ) gives ( 1 ). Differentiating ( 3y^{1/3} ) with respect to ( x ) using the chain rule gives ( \frac{1}{3} \cdot 3 \cdot y^{-2/3} \cdot \frac{dy}{dx} = y^{-2/3} \cdot \frac{dy}{dx} ). Differentiating ( 2y ) with respect to ( x ) gives ( 2 \frac{dy}{dx} ).

Putting these together, we have ( 1 = y^{-2/3} \cdot \frac{dy}{dx} + 2 \To find dy/dx, we differentiate the given equation implicitly with respect to x.

Differentiating x = 3y^(1/3) + 2y with respect to x, we get:

1 = (1/3) * 3 * y^(-2/3) * dy/dx + 2 * dy/dx

Simplifying this, we have:

1 = y^(-2/3) * dy/dx + 2 * dy/dx

Multiplying throughout by y^(2/3), we get:

y^(2/3) = dy/dx + 2y^(2/3) * dy/dx

Now, combining like terms, we get:

y^(2/3) = (1 + 2y^(2/3)) * dy/dx

So, finally, we can solve for dy/dx:

dyTo find ( \frac{dy}{dx} ) given ( x = 3y^{1/3} + 2y ), differentiate both sides of the equation with respect to ( x ) using implicit differentiation.

Differentiating ( x ) with respect to ( x ) gives ( 1 ). Differentiating ( 3y^{1/3} ) with respect to ( x ) using the chain rule gives ( \frac{1}{3} \cdot 3 \cdot y^{-2/3} \cdot \frac{dy}{dx} = y^{-2/3} \cdot \frac{dy}{dx} ). Differentiating ( 2y ) with respect to ( x ) gives ( 2 \frac{dy}{dx} ).

Putting these together, we have ( 1 = y^{-2/3} \cdot \frac{dy}{dx} + 2 \frac{dy}{dx} ).

Solving for ( \frac{dy}{dxTo find dy/dx, we differentiate the given equation implicitly with respect to x.

Differentiating x = 3y^(1/3) + 2y with respect to x, we get:

1 = (1/3) * 3 * y^(-2/3) * dy/dx + 2 * dy/dx

Simplifying this, we have:

1 = y^(-2/3) * dy/dx + 2 * dy/dx

Multiplying throughout by y^(2/3), we get:

y^(2/3) = dy/dx + 2y^(2/3) * dy/dx

Now, combining like terms, we get:

y^(2/3) = (1 + 2y^(2/3)) * dy/dx

So, finally, we can solve for dy/dx:

dy/dx = y^(2/3) / (1 + 2y^(To find ( \frac{dy}{dx} ) given ( x = 3y^{1/3} + 2y ), differentiate both sides of the equation with respect to ( x ) using implicit differentiation.

Differentiating ( x ) with respect to ( x ) gives ( 1 ). Differentiating ( 3y^{1/3} ) with respect to ( x ) using the chain rule gives ( \frac{1}{3} \cdot 3 \cdot y^{-2/3} \cdot \frac{dy}{dx} = y^{-2/3} \cdot \frac{dy}{dx} ). Differentiating ( 2y ) with respect to ( x ) gives ( 2 \frac{dy}{dx} ).

Putting these together, we have ( 1 = y^{-2/3} \cdot \frac{dy}{dx} + 2 \frac{dy}{dx} ).

Solving for ( \frac{dy}{dx}To find dy/dx, we differentiate the given equation implicitly with respect to x.

Differentiating x = 3y^(1/3) + 2y with respect to x, we get:

1 = (1/3) * 3 * y^(-2/3) * dy/dx + 2 * dy/dx

Simplifying this, we have:

1 = y^(-2/3) * dy/dx + 2 * dy/dx

Multiplying throughout by y^(2/3), we get:

y^(2/3) = dy/dx + 2y^(2/3) * dy/dx

Now, combining like terms, we get:

y^(2/3) = (1 + 2y^(2/3)) * dy/dx

So, finally, we can solve for dy/dx:

dy/dx = y^(2/3) / (1 + 2y^(2To find ( \frac{dy}{dx} ) given ( x = 3y^{1/3} + 2y ), differentiate both sides of the equation with respect to ( x ) using implicit differentiation.

Differentiating ( x ) with respect to ( x ) gives ( 1 ). Differentiating ( 3y^{1/3} ) with respect to ( x ) using the chain rule gives ( \frac{1}{3} \cdot 3 \cdot y^{-2/3} \cdot \frac{dy}{dx} = y^{-2/3} \cdot \frac{dy}{dx} ). Differentiating ( 2y ) with respect to ( x ) gives ( 2 \frac{dy}{dx} ).

Putting these together, we have ( 1 = y^{-2/3} \cdot \frac{dy}{dx} + 2 \frac{dy}{dx} ).

Solving for ( \frac{dy}{dx} \To find dy/dx, we differentiate the given equation implicitly with respect to x.

Differentiating x = 3y^(1/3) + 2y with respect to x, we get:

1 = (1/3) * 3 * y^(-2/3) * dy/dx + 2 * dy/dx

Simplifying this, we have:

1 = y^(-2/3) * dy/dx + 2 * dy/dx

Multiplying throughout by y^(2/3), we get:

y^(2/3) = dy/dx + 2y^(2/3) * dy/dx

Now, combining like terms, we get:

y^(2/3) = (1 + 2y^(2/3)) * dy/dx

So, finally, we can solve for dy/dx:

dy/dx = y^(2/3) / (1 + 2y^(2/To find ( \frac{dy}{dx} ) given ( x = 3y^{1/3} + 2y ), differentiate both sides of the equation with respect to ( x ) using implicit differentiation.

Differentiating ( x ) with respect to ( x ) gives ( 1 ). Differentiating ( 3y^{1/3} ) with respect to ( x ) using the chain rule gives ( \frac{1}{3} \cdot 3 \cdot y^{-2/3} \cdot \frac{dy}{dx} = y^{-2/3} \cdot \frac{dy}{dx} ). Differentiating ( 2y ) with respect to ( x ) gives ( 2 \frac{dy}{dx} ).

Putting these together, we have ( 1 = y^{-2/3} \cdot \frac{dy}{dx} + 2 \frac{dy}{dx} ).

Solving for ( \frac{dy}{dx} ), weTo find dy/dx, we differentiate the given equation implicitly with respect to x.

Differentiating x = 3y^(1/3) + 2y with respect to x, we get:

1 = (1/3) * 3 * y^(-2/3) * dy/dx + 2 * dy/dx

Simplifying this, we have:

1 = y^(-2/3) * dy/dx + 2 * dy/dx

Multiplying throughout by y^(2/3), we get:

y^(2/3) = dy/dx + 2y^(2/3) * dy/dx

Now, combining like terms, we get:

y^(2/3) = (1 + 2y^(2/3)) * dy/dx

So, finally, we can solve for dy/dx:

dy/dx = y^(2/3) / (1 + 2y^(2/3To find ( \frac{dy}{dx} ) given ( x = 3y^{1/3} + 2y ), differentiate both sides of the equation with respect to ( x ) using implicit differentiation.

Differentiating ( x ) with respect to ( x ) gives ( 1 ). Differentiating ( 3y^{1/3} ) with respect to ( x ) using the chain rule gives ( \frac{1}{3} \cdot 3 \cdot y^{-2/3} \cdot \frac{dy}{dx} = y^{-2/3} \cdot \frac{dy}{dx} ). Differentiating ( 2y ) with respect to ( x ) gives ( 2 \frac{dy}{dx} ).

Putting these together, we have ( 1 = y^{-2/3} \cdot \frac{dy}{dx} + 2 \frac{dy}{dx} ).

Solving for ( \frac{dy}{dx} ), we getTo find dy/dx, we differentiate the given equation implicitly with respect to x.

Differentiating x = 3y^(1/3) + 2y with respect to x, we get:

1 = (1/3) * 3 * y^(-2/3) * dy/dx + 2 * dy/dx

Simplifying this, we have:

1 = y^(-2/3) * dy/dx + 2 * dy/dx

Multiplying throughout by y^(2/3), we get:

y^(2/3) = dy/dx + 2y^(2/3) * dy/dx

Now, combining like terms, we get:

y^(2/3) = (1 + 2y^(2/3)) * dy/dx

So, finally, we can solve for dy/dx:

dy/dx = y^(2/3) / (1 + 2y^(2/3))To find ( \frac{dy}{dx} ) given ( x = 3y^{1/3} + 2y ), differentiate both sides of the equation with respect to ( x ) using implicit differentiation.

Differentiating ( x ) with respect to ( x ) gives ( 1 ). Differentiating ( 3y^{1/3} ) with respect to ( x ) using the chain rule gives ( \frac{1}{3} \cdot 3 \cdot y^{-2/3} \cdot \frac{dy}{dx} = y^{-2/3} \cdot \frac{dy}{dx} ). Differentiating ( 2y ) with respect to ( x ) gives ( 2 \frac{dy}{dx} ).

Putting these together, we have ( 1 = y^{-2/3} \cdot \frac{dy}{dx} + 2 \frac{dy}{dx} ).

Solving for ( \frac{dy}{dx} ), we get (To find dy/dx, we differentiate the given equation implicitly with respect to x.

Differentiating x = 3y^(1/3) + 2y with respect to x, we get:

1 = (1/3) * 3 * y^(-2/3) * dy/dx + 2 * dy/dx

Simplifying this, we have:

1 = y^(-2/3) * dy/dx + 2 * dy/dx

Multiplying throughout by y^(2/3), we get:

y^(2/3) = dy/dx + 2y^(2/3) * dy/dx

Now, combining like terms, we get:

y^(2/3) = (1 + 2y^(2/3)) * dy/dx

So, finally, we can solve for dy/dx:

dy/dx = y^(2/3) / (1 + 2y^(2/3))To find ( \frac{dy}{dx} ) given ( x = 3y^{1/3} + 2y ), differentiate both sides of the equation with respect to ( x ) using implicit differentiation.

Differentiating ( x ) with respect to ( x ) gives ( 1 ). Differentiating ( 3y^{1/3} ) with respect to ( x ) using the chain rule gives ( \frac{1}{3} \cdot 3 \cdot y^{-2/3} \cdot \frac{dy}{dx} = y^{-2/3} \cdot \frac{dy}{dx} ). Differentiating ( 2y ) with respect to ( x ) gives ( 2 \frac{dy}{dx} ).

Putting these together, we have ( 1 = y^{-2/3} \cdot \frac{dy}{dx} + 2 \frac{dy}{dx} ).

Solving for ( \frac{dy}{dx} ), we get ( \frac{dy}{dx} = \frac{1}{y^{-2/3} + 2} ).