Answer
$$\frac{\partial f(x,y)}{\partial x}=12x^2y^{-2}$$
$$\frac{\partial f(x,y)}{\partial y}=-8x^3y^{-3}$$
Work Step by Step
The partial derivative with the respect to $x$ is:
$$\frac{\partial f(x,y)}{\partial x}=\frac{\partial}{\partial x}(4x^3y^{-2})=4y^{-2}\frac{\partial }{\partial x}(x^3)=4y^{-2}3x^2=12x^2y^{-2}$$
because $y$ is treated as a constant.
The partial derivative with the respect to $y$ is:
$$\frac{\partial f(x,y)}{\partial y}=\frac{\partial }{\partial y}(4x^3y^{-2})=4x^3\frac{\partial }{\partial y}(y^{-2})=4x^3(-2y^{-3})=-8x^3y^{-3}$$
because now $x$ is treated as a constant.
