Answer
(a) We can see a sketch of $f(x)$ below.
(b) $f'(x)$ is differentiable for all $x$
(c) $f'(x) = 2\vert x \vert$ for all $x$
Work Step by Step
(a) $f(x) = x~\vert x \vert$
Then:
$f(x) = -x^2~~~~$ if $x \lt 0$
$f(x) = x^2~~~~$ if $x \geq 0$
We can see a sketch of $f(x)$ below.
(b) Suppose $a \lt 0$:
$f'(a) = \lim\limits_{x \to a}\frac{x\vert x \vert-a\vert a \vert}{x-a}$
$= \lim\limits_{x \to a}\frac{x(-x)-a(-a)}{x-a}$
$= \lim\limits_{x \to a}\frac{a^2-x^2}{x-a}$
$= \lim\limits_{x \to a}\frac{(a-x)(a+x)}{x-a}$
$= \lim\limits_{x \to a}(-1)(a+x)$
$= (-1)(2a)$
$ = -2a$
$ = 2 \vert a \vert$
Suppose $a \gt 0$:
$f'(a) = \lim\limits_{x \to a}\frac{x\vert x \vert-a\vert a \vert}{x-a}$
$= \lim\limits_{x \to a}\frac{x(x)-a(a)}{x-a}$
$= \lim\limits_{x \to a}\frac{x^2-a^2}{x-a}$
$= \lim\limits_{x \to a}\frac{(x-a)(x+a)}{x-a}$
$= \lim\limits_{x \to a}(x+a)$
$= 2a$
$ = 2 \vert a \vert$
Suppose $a = 0$:
$f'(a) = \lim\limits_{x \to a}\frac{x\vert x \vert-a\vert a \vert}{x-a}$
$= \lim\limits_{x \to 0}\frac{x\vert x \vert-0\vert 0 \vert}{x-0}$
$= \lim\limits_{x \to 0}\frac{x\vert x \vert}{x}$
$= \lim\limits_{x \to 0}\vert x \vert$
$= 0$
$f'(x)$ is differentiable for all $x$
(c) As shown above, $f'(x) = 2\vert x \vert$ for all $x$
