#### Answer

The proof is detailed in the Work step by step below.

#### Work Step by Step

a) $|f(x)|\le x^2$ for $-1\le x\le 1$.
Consider the inequality: $$|f(x)|\le x^2$$ $$-x^2\le f(x)\le x^2$$
Since $\lim_{x\to0}(-x^2)=\lim_{x\to0}(x^2)=0$, according to the Squeeze Theorem, $\lim_{x\to0}f(x)=0$.
Also, since $-x^2\le f(x)\le x^2$, for $x=0$: $$-0^2\le f(x)\le0^2$$ $$0\le f(x)\le 0$$
So, $f(x)=0$ for $x=0$.
We have the derivative $f'(0)$: $$f'(0)=\lim_{h\to0}\frac{f(h)-f(0)}{h}=\frac{\lim_{h\to0}f(h)-0}{h}=\frac{\lim_{h\to0}f(h)}{h}$$
Recall that $\lim_{x\to0}f(x)=0$ with $x=h$ here: $$f'(0)=\frac{0}{h}=0$$
So $f'(0)=0$. $f$ is differentiable at $x=0$.
b) $f(x)=x^2\sin1/x$ for $x\ne0$ and $f(x)=0$ for $x=0$
We know that $$-1\le\sin\frac{1}{x}\le1$$ $$-x^2\le x^2\sin\frac{1}{x}\le x^2$$ $$-x^2\le f(x)\le x^2$$ $$|f(x)|\le x^2$$
As proved in a), $f$ is differentiable at $x=0$ and $f'(0)=0$.