Let $f\left(x\right)=\left|\right|x-1|-1|$ then all the number of points where$f\left(x\right)$ is not differentiable is (are)

$\begin{array}{l}f\left(x\right)=\left\{\begin{array}{ll}|x-1|-1& \text{ if }|x-1|\ge 1\\ 1-|x-1|& \text{ if }|x-1|<1\end{array}\right.\\ =\left\{\begin{array}{cc}-x& x\le 0\\ x& 0<x\le 1\\ 2-x& 1<x<2\\ x-2& x\ge 2\end{array}\right.\end{array}$

$f$ is not differentiable at  $0, 1, 2$ as these are corner points on the graph.