Let $f : \mathbb { R } \rightarrow \mathbb { R }$ be a function which is differentiable at 0. Define another function $g : \mathbb { R } \rightarrow \mathbb { R }$ as follows:
$$g ( x ) = \begin{cases} f ( x ) \sin \left( \frac { 1 } { x } \right) & \text { if } x \neq 0 \\ 0 & \text { if } x = 0 \end{cases}$$
Suppose that $g$ is also differentiable at 0. Prove that
$$g ^ { \prime } ( 0 ) = f ^ { \prime } ( 0 ) = f ( 0 ) = g ( 0 ) = 0$$