Solution: Note that for $x\neq 0$, \[ \vert f(x) \vert = \left\vert x^2 \sin \frac{1}{x} \right\vert \leq \vert x^2 \vert. \] So the function $f$ satisfies the property that $\vert f(x) \vert \leq x^2$ in some neighborhood of zero and $f(0) = 0$. Then we claim that the function is differentiable at $x=0$. The difference quotient at $x_0 = 0$ will be \[ \frac{f(x) - f(0)}{x - 0} = \frac{f(x)}{x}, \] from which we have \begin{align*} \left\vert \frac{f(x)}{x} \right\vert \leq \left\vert \frac{x^2}{x} \right\vert = \vert x \vert . \end{align*} So, \[ 0 \leq \lim_{x \to 0} \left\vert \frac{f(x)}{x} \right\vert \leq \lim_{x \to 0} \vert x \vert = 0. \] Therefore, $f^\prime (0) = 0$. So this proves that the function is differentiable at $x = 0$.

Now we will prove that the derivative of $f$ is not continuous at $x = 0$. Just a simple computation gives us \[ f^\prime (x) = \begin{cases} \cos \frac{1}{x} + 2x \sin \frac{1}{x}, &\text{ if } x \neq 0 ;\\ 0 , &\text{ otherwise} . \end{cases} \] To prove that the $f^\prime (x)$ is not continuous at $x= 0$, we consider the following sequence: \[ x_n = \frac{1}{\pi n}. \] Clearly, $x_n \rightarrow 0$. Note that \[ f^\prime \left( x_n \right) = \begin{cases} \cos \pi n, &\text{ if } x \neq 0 ;\\ 0 , &\text{ otherwise} . \end{cases} \] Since $\cos (\pi n)$ takes the number $\pm 1$ and hence the sequence is an alternating sequence with values $1$ and $-1$. This implies $f^\prime \left( x_n \right) $ is not a convergent sequence and hence $f^\prime (x)$ is not continuous at $x = 0$.