Answer
$ a.\quad$ No.
$ b.\quad$ No.
$ c.\quad$ No.
$ d.\quad$ f(x) has absolute minimum values of 0 at $x=-3,0,3$,
and local maxima of $6\sqrt{3}$ at $x=\pm\sqrt{3}$
Work Step by Step
The absolute value function is defined as $|x|=\left\{\begin{array}{ll}
x, & x\geq 0\\
-x & x\leq 0
\end{array}\right.$.
$x^{3}-9x=x(x^{2}-9)=x(x-3)(x+3)$
We find out where this is positive:
$\left[\begin{array}{llllllllll}
& -\infty & & -3 & & 0 & & 3 & & \infty\\
x & & - & & - & & + & & + & \\
x+3 & & - & & + & & + & & + & \\
x-3 & & - & & - & & - & & + & \\
\hline & & & & & & & & & \\
x^{3}-9x & & - & & + & & - & & + &
\end{array}\right]$
So, $f(x)=\left\{\begin{array}{lll}
x^{3}-9x & for & x\in(-3,0)\cup(3,+\infty)\\
-(x^{3}-9x) & for & x\in(-\infty,-3)\cup(0,3)
\end{array}\right.$
$f'(x)=\left\{\begin{array}{lll}
3x^{2}-9 & for & x\in(-3,0)\cup(3,+\infty)\\
-3x^{2}+9 & for & x\in(-\infty,-3)\cup(0,3)
\end{array}\right.$
$(a)$
$\displaystyle \lim_{x\rightarrow 0^{-}}f'(x)=-9,\qquad \lim_{x\rightarrow 0^{+}}f'(x)=9$,
so $f'(0)$ is not defined.
$(b)$
$\displaystyle \lim_{x\rightarrow-3^{-}}f'(x)=-27+9=-18,\qquad \lim_{x\rightarrow-3^{+}}f'(x)=27-9=18$,
so $f'(-3)$ is not defined.
$(c)$
$\displaystyle \lim_{x\rightarrow 3^{-}}f'(x)=27-9=18,\qquad \lim_{x\rightarrow 3^{+}}f'(x)=-27+9=-18$,
so $f'(3)$ is not defined.
$(d)$
Critical points:
$-3,0,+3$ ($f'$ is undefined)
and
$\pm\sqrt{3}\quad$ ($f'(x)=0$)
$f(0)=f(3)=f(-3)=0,$
$f(-\sqrt{3})=|-3\sqrt{3}+9\sqrt{3})=6\sqrt{3}$
$f(\sqrt{3})=|3\sqrt{3}-9\sqrt{3})=6\sqrt{3}$
When $ x\rightarrow\pm\infty,\quad f(x)\rightarrow+\infty$.
So, f(x) has absolute minimum values of 0 at $x=-3,0,3$,
and local maxima of $6\sqrt{3}$ at $x=\pm\sqrt{3}$
