Answer
$x=-2$ is a local maximum and $x=0$ is a local minimum.
Work Step by Step
Given $$f(x)=\frac{x^{2}}{x+1} $$
Since
\begin{align*}
f'(x)&= \frac{\frac{d}{dx}\left(x^2\right)\left(x+1\right)-\frac{d}{dx}\left(x+1\right)x^2}{\left(x+1\right)^2}\\
&=\frac{x^2+2x}{\left(x+1\right)^2}
\end{align*}
Then $f'(x)=0$ for $x=0$ and $x=-2$ and $f'(x)$ is not defined at $x=-1$. To determine the local maximum and local minimum, choose $x= -3$, $x= -1.5$, $x=-0.5$ and $x=1$:
\begin{align*}
f'( -3 )&>0 \\
f'(-1.5)&<0 \\
f'(-0.5)&<0 \\
f'(1)&>0
\end{align*}
Then, by using the First Derivative Test, we see that $x=-2$ is a local maximum and $x=0$ is a local minimum.
