Answer
$$\frac{dy}{dx}=\frac{3x^2y^2-4x^3}{\cos y-2x^3y}$$
Work Step by Step
$$x^4+\sin y=x^3y^2$$
1) Differentiate both sides of the equation with respect to $x$:
$$\frac{d}{dx}(x^4)+\frac{d}{dx}(\sin y)=\frac{d}{dx}(x^3y^2)$$
$$4x^3+\cos y\frac{dy}{dx}=3x^2y^2+x^3(2y)\frac{dy}{dx}$$
$$4x^3+\cos y\frac{dy}{dx}=3x^2y^2+2x^3y\frac{dy}{dx}$$
2) Collect all the terms with $dy/dx$ onto one side and solve for $dy/dx$:
$$\cos y\frac{dy}{dx}-2x^3y\frac{dy}{dx}=3x^2y^2-4x^3$$
$$\frac{dy}{dx}=\frac{3x^2y^2-4x^3}{\cos y-2x^3y}$$
