If $f\left(x\right)=\left\{\begin{array}{cc}{x}^2\cos \left(\frac{1}{x}\right)& \text{ for }x\ne 0\\ 0& \text{ for }x=0\end{array}\right.$, then 

$x\ne 0$

$\begin{array}{l}\Rightarrow f^1\left(x\right)=2x\cos \left(\frac{1}{x}\right)+x^2\cdot \sin \left(\frac{1}{x}\right)\frac{1}{x^2}\\ =2x\cos \left(\frac{1}{x}\right)+\sin \left(\frac{1}{x}\right)\\ {\mathrm{Lt}}_{x\to 0}\frac{f\left(x\right)-f\left(0\right)}{x-0}{=}_{x\to 0}^{Lt}x\cos \left(\frac{1}{x}\right)=0\\ f^1\left(0\right)=0\end{array}$

$f\text{ is derivable at }x=0$

${\mathrm{Lt}}_{x\to 0}{f}^1\left(x\right)={\mathrm{Lt}}_{x\to 0}2x\cos \left(\frac{1}{x}\right)+\sin (1/x)$

Does not exist

$f^1\text{ is not continuous at }x=0$