Let f(x) and g(x) be two real functions such that f(x)=x/(|x|) g(x),…

We have,

$f(x)=\frac{x}{|x|} g(x), x \neq 0 $

$=\left\{\begin{array}{rr}g(x), & x>0 \\ -g(x), & x<0\end{array}\right.$

It is given that g'(0) = g(0) = 0. This means that g(x) is differentiable and hence continuous at x = 0 and $\lim\limits_{x \rightarrow 0} g(x)=g(0)=0$

Now,

$\lim\limits_{x \rightarrow 0^{-}} f(x)=\lim\limits_{x \rightarrow 0^{-}}-g(x)=-\lim\limits_{x \rightarrow 0} g(x)=0$

and,

$\lim\limits_{x \rightarrow 0^{+}} f(x)=\lim\limits_{x \rightarrow 0^{+}} g(x)=\lim\limits_{x \rightarrow 0} g(x)=0$

∴  $\lim\limits_{x \rightarrow 0^{-}} f(x)=\lim\limits_{x \rightarrow 0^{+}} f(x)$

So, f(x) is continuous at x = 0.

∴  $\lim\limits_{x \rightarrow 0} f(x)=f(0)$

$\Rightarrow f(0)=0$

Thus, we have

$f(x)=\left\{\begin{array}{rr}g(x), & x>0 \\ 0, & x=0 \\ -g(x), & x<0 \end{array}\right.$

Now,

(LHD of f(x) at x = 0)

$=\lim\limits_{h \rightarrow 0} \frac{f(0-h)-f(0)}{-h}$

$=\lim\limits_{h \rightarrow 0} \frac{f(-h)}{-h}$

$=\lim\limits_{h \rightarrow 0} \frac{-g(-h)}{-h}$

$=\lim\limits_{h \rightarrow 0} \frac{g(-h)-g(0)}{-h}$               [∵  g(0) = 0]

$=-g'(0)=0$

Similarly, we have

(RHD of f(x) at x = 0) = g'(0) = 0.

Hence, f(x) is differentiable at x = 0 and f'(0) =0.