$f\left(x\right)=\left\{\begin{array}{ccc}x\sin 1/x& ,& x\ne 0\\ 0& ,& x=0\end{array}\text{ at }x=0\right.$ is.

For function to be continuous 

$\begin{array}{l}f(0+h)=f(0-h)=f\left(0\right)\\ f(0+h)=\underset{h\to 0}{lim} h\sin 1/h=0\times \left(\text{ a finite quantity }\right)=0\end{array}$

$\mathrm{f}(0-\mathrm{h})=\underset{\mathrm{h}\to 0}{lim} -\mathrm{hsin}(\mathrm{l}/\mathrm{h})=0\times \left(\text{ a finite quantity }\right)=0$

Also, $\underset{x\to 0}{lim} x\sin 1/x=0\times \left(\text{ a finite quantity }\right)=0$

$\Rightarrow$function is continuous at $x=0$

For function to be differentiable :

$\begin{array}{l}{f}^{\mathrm{\prime }}(0+h)=f^{\mathrm{\prime }}(0-h)\\ f^{\mathrm{\prime }}(0+h)=\underset{h\to 0}{lim}\frac{f(0+h)-f\left(0\right)}{h}\\ =\underset{h\to 0}{lim} \frac{h\sin \frac{1}{h}-0}{h}=\underset{h\to 0}{lim} \sin \left(\frac{1}{h}\right)\end{array}$

which does not exist.

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

which does not exist. So function is not differentiable at $x=0$