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