#### Answer

$f(x)$ has a local minimum at $x=1$ and a local maximum at $x=-1$

#### Work Step by Step

Given $$f(x)=x^{3}+x^{-3}$$
Since
$$f'(x) =3 x^{2}-3 x^{-4} $$
Then $f(x)$ has critical points when
\begin{align*}
f'(x)&=0\\
3 x^{2}-3 x^{-4} &=0\\
\frac{3}{x^{4}}\left(x^{6}-1\right)&=0\\
x^{6}-1&=0
\end{align*}
Then $x= -1$ and $x=1$ are critical points. To find the intervals where the function is increasing and decreasing, choose $x=-2,\ x= 0$ and $x=2$
\begin{align*}
f'( -2)&>0 \\
f'(0)& <0\\
f'(2)&>0
\end{align*}
Hence, by using the first derivative test, $f(x)$ has a local minimum at $x=1$ and a local maximum at $x=-1$.