$\left|g\left(x\right)\right|=\left\{\begin{array}{ll}\left|-1\right|& -2\le x\le 0\\ \left|x^2-1\right|& 0\le x\le 2\end{array}\right.$

             $=\left\{\begin{array}{ll}1& -2\le x\le 0\\ 1-x^2& 0\le x<1\\ x^2-1& 1\le x\le 2\end{array}\right.$

And $g\left(\left|x\right|\right)=\left\{\begin{array}{ll}-1& -2\le \left|x\right|<0\\ x^2-1& 0\le \left|x\right|\le 2\end{array}\right.$

So, $g\left(\left|x\right|\right)=x^2-1, -2\le x\le 2$

Hence,   $f\left(x\right)=\left\{\begin{array}{ll}{x}^2& -2\le x<0\\ 0& 0\le x<1\\ 2\left(x^2-1\right)& 1\le x\le 2\end{array}\right.$

$f^{\text{'}}\left(x\right)=\left\{\begin{array}{ll}2x,& -2<x<0\\ 0,& 0<x<1\\ 4x,& 1<x<2\end{array}\right.$

This clearly shows $f\left(x\right)$ is not differentiable at $x=1$.