The function is $$f(x)= \begin{cases}2x^4 +x^4\sin(1/x),x\neq0 \\ 0,x=0 \end{cases}$$
To show differentiability when $x\neq 0$, it should be sufficient to compute the derivative with the product and chain rule. The result is $$f'(x)=(x^2)(8x-\cos (1/x)+4x\sin(1/x))$$
, which is not defined at $x=0$. Therefore we use the definition of a derivative and gain the quotient $$\phi(h)=\frac{f(h)-f(0)}{h}=2h^3+h^3\sin(1/h).$$
This is where I don't quite understand the logic. I believe we can show that ( by the squeeze theorem? ) $$ 0 \leq\vert 2h^3+h^3\sin(1/h)\vert\leq \vert2h^3+h^3\vert$$
, and if $h\to 0$ then $\phi(h) \to 0$. But isn't $\sin(1/h)$ still a problem, even if $h\neq 0$?
No, it is not. We have$$\left\lvert2h^3+h^3\sin\left(\frac1h\right)\right\rvert=\lvert h\rvert^3\left\lvert2+\sin\left(\frac1h\right)\right\rvert\leqslant3\lvert h\rvert^3$$for every $h\neq0$, since $\left\lvert\sin\left(\frac1h\right)\right\rvert\leqslant1$. So, yes, you can apply the squeeze theorem here.