Answer
$\frac{dy}{dx}=\frac{2-3x^2-4y}{4x-4y^{1/3}}$
Work Step by Step
Take the derivative of the equation on each side separately. Apply chain rule when differentiating the "y" variables since we are differentiating with respect to x:
$3x^2+4x\frac{dy}{dx}+4y-\frac{4}{3}\times3y^{1/3}\frac{dy}{dx}=2$
Move all terms with dy/dx to one side of the equation, and isolate dy/dx:
$4x\frac{dy}{dx}-4y^{1/3}\frac{dy}{dx}=2-3x^2-4y$
$\frac{dy}{dx}(4x-4y^{1/3})=2-3x^2-4y$
$\frac{dy}{dx}=\frac{2-3x^2-4y}{4x-4y^{1/3}}$
