Answer
a) $(-\infty,-3)\cup (-3,3)\cup (3,\infty)$
b) See graphs
Work Step by Step
a) Rewrite the function:
$$\begin{align*}
f(x)=\begin{cases}
x^2-9,&& x<-3\\
-x^2+9,&&-3\leq x\leq 3\\
x^2-9,&&x>3
\end{cases}
\end{align*}$$
Determine the first derivative $f'(x)$:
$$\begin{align*}
f'(x)=\begin{cases}
2x,&& x<-3\\
-2x,&&-33
\end{cases}
\end{align*}$$
We notice that at $x=-3$ and $x=-3$, $f$ is not differentiable.
b) Graph $f$ and $f'$.
