We have, $f(x)=\left\{\begin{array}{cl}x^{2}, & x \geq 0 \\ -x^{2}, & x<0\end{array}\right.$
Clearly, $f(x)$ is differentiable for all $x>0$ and for all $x<0.$
So, we check the differentiability at $x=0$
Now, (RHD at $x=a$ )
$=\left(\frac{d}{d x}\left(x^{2}\right)\right)_{x=0}=(2 x)_{x=0}=0$
$\therefore ( LHD \text { at } x=0)=\left(\frac{d}{d x}\left(-x^{2}\right)\right)_{x=0}=(-2 x)_{x=0}=0$
$( LHD$ at $x=0)=( RHD$ at $x=0)$
So, $f(x)$ is differentiable for all $x$ ie, the set of all points where $f(x)$ is differentiable is $(-\infty, \infty) i e,R$