Answer
$f(x)$ increasing on $\left( -\infty,-3\right )\cup (3,\infty)$ and decreasing on $ (-3,-\sqrt{3})\cup ( 0,\sqrt{3})\cup (\sqrt{3}, 3)$
$f(x)$ has a local maximum at $x=-3$ and a local minimum at $x=3$.
Work Step by Step
Given $$y=\frac{x^{3}}{x^{2}-3}$$
Since
\begin{align*}
f'(x) &=\frac{\frac{d}{dx}\left(x^3\right)\left(x^2-3\right)-\frac{d}{dx}\left(x^2-3\right)x^3}{\left(x^2-3\right)^2}\\
&=\frac{x^4-9x^2}{\left(x^2-3\right)^2}
\end{align*}
Then $f'(x)=0$ for $x^4-9x^2= 0 $. Hence, $$x=0,\ \ \ \ x=-3,\ \ \ \ x=3 ,\ \ \ \ x=\pm \sqrt{3}$$
Choose $x=-1,\ \ x=1 $:
\begin{align*}
f'(-4)&> 0\\
f'(-2)&<0\\
f'(-1) &<0\\
f'(2)&<0\\
f'(4)&>
\end{align*}
Then $f(x)$ is increasing on $\left( -\infty,-3\right )\cup (3,\infty)$ and decreasing on $ (-3,-\sqrt{3})\cup ( 0,\sqrt{3})\cup (\sqrt{3}, 3)$. Hence, $f(x)$ has a local maximum at $x=-3$ and a local minimum at $x=3$.
