I'm struggling to prove from first principles that the function $f(x)$ is differentiable.
Definition of differentiable:
Let $S$ be an open interval and let $f : S \rightarrow \Bbb R$ be a function. Let $x_0 \in S$. we say that $f$ is differentiable at $x_0$ if the limit $$\lim_{x\to x_0} \frac {f(x)-f(x_0)} {x-x_0}$$ exists. If this is the case, then the limit i denoted by $f'(x_0)$ and called the derivative of $f$ at $x_0$.
Prove from first principles that the function $g : \Bbb R \rightarrow \Bbb R$ defined by $$g(x) = \begin{cases} 0, & 0 \\ x^3\sin\left(x^{-4}\right), & x \ne 0 \end{cases} $$ is differentiable.