#### Answer

$f(x)$ increasing on $\left( -\infty,0\right )\cup (0,\infty)$.
$f(x)$ has neither a local maximum nor a local minimum.

#### Work Step by Step

Given $$y =\frac{x^{3}}{x^{2}+1}$$
Since
\begin{align*}
f'(x) &=\frac{\frac{d}{dx}\left(x^3\right)\left(x^2+1\right)-\frac{d}{dx}\left(x^2+1\right)x^3}{\left(x^2+1\right)^2}\\
&= \frac{x^4+3x^2}{\left(x^2+1\right)^2}
\end{align*}
Then $f'(x)=0$ for $x^4+3x^2= 0 $. Hence, $x=0 $.
Choose $x=-1,\ \ x=1 $. Then:
\begin{align*}
f'(-1)&> 0\\
f'(1)&>0
\end{align*}
Thus $f(x)$ is increasing on $\left( -\infty,0\right )\cup (0,\infty)$ and hence $f(x)$ has neither a local maximum nor a local minimum.