$f(0)=0$ and $\lim_{x \to \infty}f(x)=0$ and $f$ is differentiable in $\mathbb{R}$
How can I show there's $x_0$ such that $f'(x_0)=0$.
I know it's true since the values of f(x) for large enough x are close to 0 as much as I want. So, that means that the same value would have to appear at least twice, and then it can be concluded. However, I don't know how to show it formally.
If $f$ is identically zero there isn't much to show.
If $f(x_1) > 0$ for some $x_1$, you can use the limit to deduce there is $M > 0$ for which $$x \ge M \implies |f(x)| < \dfrac{f(x_1)}{2}.$$
The function $f$ has a maximum on the interval $[0,M]$, and it doesn't occur at either endpoint since $f(x_1)$ exceeds both $f(0)$ and $f(M)$.
The derivative vanishes at this maximum.
If $f(x_1) < 0$ for some $x_1$ you can argue similarly.