Let $f\left(x\right)=\left\{\begin{array}{ll}{x}^2\sin \left(\frac{1}{x}\right),& x\ne 0\\ 0,& x=0\end{array}\right.\begin{array}{} \\ \end{array}$ ; Then at  $x=0$

$-1\le \sin \left(\frac{1}{x}\right)\le 1\text{⇒}-x^2\le x^2\sin \left(\frac{{\displaystyle 1}}{{\displaystyle x}}\right)\le x^2$

$$\begin{gathered} \underset{x\to 0^{+}}{lim} f\left(x\right)=0=\underset{x\to 0^{-}}{lim} f\left(x\right)=f\left(0\right) \\ \text{ Therefore, }f\left(x\right)\text{ is continuous at }x=0\text{. } \end{gathered}$$

Since $f\text{'}\left(x\right)=2x.\sin \left(\frac{1}{x}\right)+x^2.\cos \left(\frac{1}{x}\right).\frac{-1}{x^2}$

so, $f\text{'}\left(x\right)=\left\{\begin{array}{cc}2x.\sin \left(\frac{1}{x}\right)-\cos \left(\frac{1}{x}\right);& x\ne 0\\ 0,& x=0\end{array}\right\}$

So, $f\left(x\right)$ is continuous and differentiable and  $f\text{'}\left(x\right)$ is not  continuous (as $\left(\frac{1}{x}\right)$ is highly oscillating at $x=0$