Answer
$$y^2,\ \ 2xy$$
Work Step by Step
Since $$\frac{\partial}{\partial x} f(x,y)= \lim _{h \rightarrow 0} \frac{f(x+h,y) -f(x ,y)}{h} $$
Then
\begin{aligned}
\frac{\partial}{\partial x} x y^{2} &=\lim _{h \rightarrow 0} \frac{(x+h) y^{2}-x y^{2}}{h}\\
&=\lim _{h \rightarrow 0} \frac{x y^{2}+h y^{2}-x y^{2}}{h}\\
&=\lim _{h \rightarrow 0} \frac{h y^{2}}{h}\\
&=\lim _{h \rightarrow 0} y^{2}\\
&=y^{2} \\
\end{aligned}
and $$\frac{\partial}{\partial y} f(x,y)= \lim _{k \rightarrow 0} \frac{f(x,y+k) -f(x ,y)}{k} $$
Then
\begin{aligned}
\frac{\partial}{\partial y} x y^{2} &=\lim _{k \rightarrow 0} \frac{x(y+k)^{2}-x y^{2}}{k}\\
&=\lim _{k \rightarrow 0} \frac{x\left(y^{2}+2 y k+k^{2}\right)-x y^{2}}{k}\\
&=\lim _{k \rightarrow 0} \frac{x y^{2}+2 x y k+x k^{2}-x y^{2}}{k} \\
&=\lim _{k \rightarrow 0} \frac{k(2 x y+x k)}{k}\\
&=\lim _{k \rightarrow 0}(2 x y+k)=2 x y+0\\
&=2 x y
\end{aligned}
