Answer
$f_{x}(x, y)=y^{2}-3 x^{2} y$
$f_{y}(x, y)=2 x y-x^{3}$
Work Step by Step
$$
f(x,y)=xy^{2}-x^{3}y
$$
The partial derivative of $f$ with respect to $x$ is given by:
$$
\begin{aligned} f_{x}(x, y) &=\lim _{h \rightarrow 0} \frac{f(x+h, y)-f(x, y)}{h} \\
&=\lim _{h \rightarrow 0} \frac{(x+h) y^{2}-(x+h)^{3} y-\left(x y^{2}-x^{3} y\right)}{h} \\
& =\lim _{h \rightarrow 0} \frac{h\left(y^{2}-3 x^{2} y-3 x y h-y h^{2}\right)}{h}\\
&=\lim _{h \rightarrow 0}\left(y^{2}-3 x^{2} y-3 x y h-y h^{2}\right) \\
&=y^{2}-3 x^{2} y. \end{aligned}
$$
The partial derivative of $f$ with respect to $y$ is given by:
$$
\begin{aligned} f_{y}(x, y) &=\lim _{h \rightarrow 0} \frac{f(x, y+h)-f(x, y)}{h} \\
& =\lim _{h \rightarrow 0} \frac{x(y+h)^{2}-x^{3}(y+h)-\left(x y^{2}-x^{3} y\right)}{h}\\
&=\lim _{h \rightarrow 0} \frac{h\left(2 x y+x h-x^{3}\right)}{h} \\ &=\lim _{h \rightarrow 0}\left(2 x y+x h-x^{3}\right)\\
&=2 x y-x^{3}. \end{aligned}
$$
