#### Answer

$f(x) $ is increasing on $(-\infty,0) \cup (8,\infty) $ and decreasing on $(0,8) $
$f(x)$ has a local maximum at $x= 0$ and a local minimum at $x=8 $

#### Work Step by Step

Given $$y=x^{3}-12 x^{2}$$
Since
$$f'(x) =3x^{2}-24 x$$
Then $f(x)$ has critical points when
\begin{align*}
f'(x)&=0\\
3x^{2}-24 x&=0\\
3x(x-8)&=0
\end{align*}
Then $x= 0$ and $x= 8$ are critical points. To find the interval where $f$ is increasing and decreasing, choose $x=-1$, $x= 1$ and $x= 9$
\begin{align*}
f'(-1)&= 27>0\\
f'(1)&=-21<0\\
f'(-1)&= 27>0
\end{align*}
Hence, $f(x)$ is increasing on $(-\infty,0) \cup (8,\infty) $ and decreasing on $(0,8) $. Hence, by using the first derivative test, $f(x)$ has a local maximum at $x= 0$ and a local minimum at $x=8$.