If $f\left(x\right)=xg\left(x\right)/\left|x\right|,g\left(0\right)=g\text{'}\left(0\right)$  and $f\left(x\right)$  is continuous at $x=0$ . Then

$f\left(x\right)$ $=\frac{xg\left(x\right)}{\left|x\right|}$ $=\left\{\begin{array}{l}-g\left(x\right) x<0\\ g\left(x\right) x>0\end{array}\right.$ 

$f\left(x\right)$  is continuous at $x=0$  $\Rightarrow f\left(0\right)=g\left(0\right)=0$

$f\text{'}\left(x\right)$ $=\left\{\begin{array}{l}-g\text{'}\left(x\right):x<0\\ g\text{'}\left(x\right):x>0\end{array}\right.$

$\therefore f\text{'}\left(0\right)=g\text{'}\left(0\right)=0$ .