Answer
$$\frac{d}{dx}\frac{x^3}{y^2}=\frac{3x^2}{y^2}-\frac{2x^3}{y^3}\frac{dy}{dx}.$$
Work Step by Step
Recall the quotient rule: $(\dfrac{u}{v})'=\dfrac{vu'-uv'}{v^2}$
Recall that $(x^n)'=nx^{n-1}$
Using the quotient and chain rules, we have
$$\frac{d}{dx}\frac{x^3}{y^2}=\frac{y^2(3x^2)-x^3(2y\frac{dy}{dx})}{y^4}$$
$$=\frac{3x^2}{y^2}-\frac{2x^3}{y^3}\frac{dy}{dx}$$
