If $f\left(x\right)=|1-x|$, then the points where ${\sin }^{-1}\left(f\right(\left|x\right|\left)\right)$ is non-differentiable, are

 Here, $f\left(x\right)=|1-x|$

Taking absolute value both sides we get,

$\Rightarrow f\left(\right|x\left|\right)=|1-|x\left|\right|$

Splitting the function we get,

$\Rightarrow \mathrm{f}\left(\mathrm{x}\right)=\left\{\begin{array}{l}-1-\mathrm{x} , \mathrm{x}<-1\\ 1+\mathrm{x}, -1\le \mathrm{x}<0\\ 1-\mathrm{x} , 0\le \mathrm{x}<1\\ \mathrm{x}-1, \mathrm{x}\ge 1\end{array}\right.$

As, we know domain of ${\sin }^{-1}\left(x\right)$ is $-1\le x\le 1$,

$\Rightarrow -1\le f\left(\right|x\left|\right)\le 1$

Since $\left|1-\left|\mathrm{x}\right|\right|\ge 0 , \forall \mathrm{x}$

$\Rightarrow |1-|x\left|\right|\le 1$

$\Rightarrow -1\le 1-\left|x\right|\le 1$

$\Rightarrow -2\le -\left|x\right|\le 0$

$\Rightarrow 2\ge \left|x\right|\ge 0$

$\Rightarrow x\in [-2,2]$

Then, ${\sin }^{-1}\left(f\right(\left|x\right|\left)\right)$, for $x\in [-2,2]$ is not differentiable at points where $f\left(\left|x\right|\right)$ is non differentiable.

The graph of $f\left(\left|x\right|\right)=\left|1-\left|x\right|\right|$ is,

$f\left(\left|x\right|\right)$ in non-differentiable where it's graph is sharp and pointed  that is $x=\{0,1,-1\}$.

Hence, option-$3$ is the correct answer.