If $f\left(x\right)=\left\{\begin{array}{ll}{x}^a\sin \left(\frac{1}{x}\right)& ,x\ne 0\\ 0& ,x=0\end{array}\right.$ is continuous, but non-differentiable at $x=0$, then

$f\text{'}\left(0\right)=\underset{h\to 0}{\lim }\frac{f\left(h\right)-f\left(0\right)}{h}$$=\underset{h\to 0}{\lim }\frac{h^a\sin \left(\frac{1}{h}\right)}{h}=\underset{h\to 0}{\lim }{h}^{a-1}\sin \left(\frac{1}{h}\right)$

This limit will not exist if $a\le 1$

$\therefore \underset{x\to 0}{\lim }f\left(x\right)=\underset{x\to 0}{\lim }{x}^a\sin \left(\frac{1}{x}\right)=0$, if a>0

$\therefore a\in \left(0,1\right]$