Let $f:R\to R$ be a function defined by $f\left(x\right)=\max \left\{x,x^3\right\}.$ The set of all points where $f\left(x\right)$ is not differentiable is

If $x<-1$ then $x>x^3$. So, $f\left(x\right)=x,$ If $x=-1$ then $x=x^3$. So,  $f\left(x\right)=x$,

If $-1<x<0$ then $x<x^3$. So, $f\left(x\right)=x,$ If $x=0$ then $x=x^3$. So,  $f\left(x\right)=x^3$, If $0<x<1$, then $x>x^3$. So, $f\left(x\right)=x,$ if $x=1$ then $x=x^3$. So $f\left(x\right)=x,$ If $x>1$, then $x<x^3$. So,  $f\left(x\right)=x^3$

Thus $$\begin{gathered} f\left(x\right)=x, x\le -1, \\ f\left(x\right)=\left\{\begin{array}{l}x, x\le -1\\ x^3, -1<x\le 0\\ x, 0<x\le 1\\ x^3, x>1\end{array}\right. \end{gathered}$$

Clearly, $f\left(x\right)$ is not differentiable at $x= -1, 0, 1.$