Answer
$\frac{4x^3-3x^2y^2}{2x^3y-cos(y)} $
Work Step by Step
Given the equation $x^4+sin(y)=x^3y^2$, differentiating both sides with respect to $x$, we have
$4x^3+cos(y)\frac{dy}{dx}=3x^2y^2+2x^3y\frac{dy}{dx}$
Solve for $\frac{dy}{dx}$: $(2x^3y-cos(y))\frac{dy}{dx}=4x^3-3x^2y^2$, we get
$\frac{dy}{dx}=\frac{4x^3-3x^2y^2}{2x^3y-cos(y)} $
