We have, $\mathrm{g}\left(\mathrm{x}\right)=\left\{\begin{array}{ll}{\mathrm{x}}^3\sin \frac{1}{\mathrm{x}}& :x\ne 0\\ 0& :x=0\end{array}\right.$
For $\mathrm{x}\ne 0,$
${\text{g}}^{\prime }\left(\text{x}\right)={\text{x}}^3\cos \frac{1}{\text{x}}\left(-\frac{1}{{\text{x}}^2}\right)+3{\text{x}}^2\sin \frac{1}{\text{x}}$
$=-\mathrm{x}\cos \frac{1}{\mathrm{x}}++3{\mathrm{x}}^2\sin \frac{1}{\mathrm{x}}$
For $\mathrm{x}=0,$
${\text{g}}^{\text{'}}\left(0\right)=\underset{\text{x}\to 0}{\lim }\frac{\text{g}\left(\text{x}\right)-\text{g}\left(0\right)}{\text{x}-0}=\underset{\text{x}\to 0}{\lim }\frac{{\text{x}}^3\sin \frac{1}{\mathrm{x}}-0}{\text{x}}$
$=\underset{\text{x}\to 0}{\lim }{\mathrm{x}}^2\sin \frac{1}{\mathrm{x}}=0.$
$\therefore {\text{g}}^{\text{'}}\left(\text{x}\right)=\left\{\begin{array}{c}3{\text{x}}^2\sin \frac{1}{\text{x}}-x\cos \frac{1}{\text{x}}, x\ne 0\\ 0, x=0\end{array}\right.$
${\text{g}}^{\prime }$ is continuous at $\mathrm{x}=0$
Also, $\mathrm{g}$ is differentiable at $\mathrm{x}=0$ .