I have some troubles with the following exercise. I think that it's trivial but I can't see the solution.
Let $f,h:\mathbb{R}\to\mathbb{R}$ be continuous functions over $\mathbb{R}$. Given that for all $x\in\mathbb{R}$ we have $$0\leq f(x)\leq xh(x)$$ and that the derivative of $xh(x)$ exists at $0$ (in fact, $\big(xh(x)\big)'(0)=h(0)$). Can we conclude that $f'(0)$ exists?
Firts, because $0\leq f(0)\leq 0h(0)=0$, we have that $f(0)=0$. Next $$f'(0)=\lim\limits_{x\to 0}\dfrac{f(x)-f(0)}{x-0}=\lim\limits_{x\to 0}\dfrac{f(x)-0}{x}=\lim\limits_{x\to 0}\dfrac{f(x)}{x}.$$ By the inequality of the hypothesis, we have $0\leq \dfrac{f(x)}{x}\leq h(x)$. Thus $$0\leq \lim\limits_{x\to 0}\dfrac{f(x)}{x}\leq \lim\limits_{x\to 0} h(x)=h(0).$$ But then what can I do? If $h(0)=0$ the result follows. But if $h(0)>0$ I don't know how to proceed.
You are on the right track. It suffices to show that $h(0)=0$.
Note that $0\leq xh(x)$ for all $x\in\mathbb{R}$ implies that $h(x)\geq 0$ for $x\ge0$ and $h(x)\leq 0$ for $x\le0$. Hence, since $h$ is continuous, it follows that $h(0)=0$.