Given,

$$f(x) = \left\{ {\matrix{ {{x^2}\sin \left( {{1 \over x}} \right),} & {x \ne 0} \cr {0,} & {x = 0} \cr } } \right.$$

$$\therefore$$ $$f'(x) = 2x\sin \left( {{1 \over x}} \right) - \cos \left( {{1 \over x}} \right)$$

Now,

$$\mathop {\lim }\limits_{x \to 0} f'(x) = \mathop {\lim }\limits_{x \to 0} \left[ {2x\sin \left( {{1 \over x}} \right) - \cos \left( {{1 \over x}} \right)} \right]$$

$$ = 0 - \mathop {\lim }\limits_{x \to 0} \cos \left( {{1 \over x}} \right)$$

= Does not exist

$$\therefore$$ $$f'(x)$$ is discontinuous function at $$x = 0$$.

L.H.D. $$ = \mathop {\lim }\limits_{h \to {0^ + }} {{f(0 - h) - f(0)} \over { - h}}$$

$$ = \mathop {\lim }\limits_{h \to {0^ + }} {{f( - h)} \over { - h}}$$

$$ = \mathop {\lim }\limits_{h \to {0^ + }} {{ - {h^2}\sin \left( {{1 \over h}} \right)} \over { - h}} = 0$$

R.H.D. $$ = \mathop {\lim }\limits_{h \to {0^ + }} {{f(0 + h) - f(0)} \over h}$$

$$ = \mathop {\lim }\limits_{h \to {0^ + }} {{f(h)} \over h}$$

$$ = \mathop {\lim }\limits_{h \to {0^ + }} {{{h^2}\sin \left( {{1 \over h}} \right)} \over h} = 0$$

$$\therefore$$ L.H.D. = R.H.D.

$$ \Rightarrow f(x)$$ is differentiable at $$x = 0$$. So, f(x) is continuous.