Let $g\left(x\right)=xf\left(x\right)\text{, where }f\left(x\right)=\left\{\begin{array}{cc}x\sin \frac{1}{x},& x\ne 0\\ 0,& x=0\end{array}\text{.At }x=0\right.$

$\underset{x\to 0}{lim} f\left(x\right)=\underset{x\to 0}{lim} x\sin \left(\frac{1}{x}\right)=0\times (\text{ any value between }-1\text{ and }1)=0$

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

which is not a fixed value.

Hence, $f\left(x\right)$ is non–differentiable at $x = 0.$

$\text{ Clearly, }g\left(x\right)=xf\left(x\right)=\left\{\begin{array}{cc}{x}^2\sin \left(\frac{1}{x}\right),& x\ne 0\\ 0,& x=0\end{array}\right.$

is continuous at $x = 0.$

Also $g\left(x\right)$ is differentaible at$x = 0.( as f(0) = 0)$ with $g^{\mathrm{\prime }}\left(0\right)=0$

$\text{ If }x\ne 0,g^{\mathrm{\prime }}\left(x\right)=x^2\cos \left(\frac{1}{x}\right)\left(-\frac{1}{x^2}\right)+2x\sin \left(\frac{1}{x}\right)$

$=-\cos \left(\frac{1}{x}\right)+2x\sin \left(\frac{1}{x}\right)\text{ which exists for }\mathrm{\forall }x\ne 0$

$\text{ If }x=0\text{ then }{g}^{\mathrm{\prime }}\left(0\right)=\underset{x\to 0}{lim} \frac{x^2\sin (1/x)-0}{x-0}=\underset{x\to 0}{lim} x\sin \left(\frac{1}{x}\right)=0$

$\Rightarrow g^{\mathrm{\prime }}\left(x\right)=\left\{\begin{array}{cc}-\cos \left(\frac{1}{x}\right)+2x\sin \frac{1}{x},& x\ne 0\\ 0,& x=0\end{array}\right.$

$\text{ At }x=0,\cos \left(\frac{1}{x}\right)\text{ is not continuous, }$

therefore, $g^{\mathrm{\prime }}\left(x\right)$ is not continuous at $x = 0.$