Differential function problem Let $f: \mathbb{R} → \mathbb{R} $ be a function such that $f(x)$ is differentiable on all $\mathbb{R}$ and $\lim_{x\to \infty}(f(x)-f(-x))=0$.

Prove there exists $x_{0} \in \mathbb{R}$ such that $f'(x_{0})=0$

I tried proving it by contradiction using that the limits at infinity and minus infinity are equal I'm having hard time formalizing it.

Any help appreciated.
Either $f$ has a local extremum somewhere at $x_0\in\mathbb R$ (a maximum or minimum), either it is monotone. In the first case, then necessarily $f'(x_0)=0$, in the second case, the fact that $\lim_{x\to+\infty}(f(x)-f(-x))=0$ implies that $f$ is constant, and thus $f'=0$ everywhere.
Since $\lim_{x\to \infty}(f(x)-f(-x))=0$,

and $f(x)$ is differentiable and assumed continuous everywhere, by the Intermediate Value Theorem, there exists a $x_{0}$ such that:

$$f^{'}(x_{0})=\frac{\lim_{x\to \infty}f(x)-\lim_{x\to \infty}f(-x)}{x+x}=0.$$ (proved)