So, this should be a simple task, but I'm not able to find an argument: Assume $f\in C^2(\mathbb R)$ and $(x_0,\delta)\in\mathbb R\times(0,\infty)$ with $$f(x+h)=-f(x-h)\tag1$$ for all $(x,h)\in(x_0-\delta,x_0+\delta)\times[0,\infty)$. Why can we conclude that $f=0$?
If it doesn't follow immediately from $(1)$ by some simple argument which I don't see, I thought we may be able to show that $f'=0$ in a first step by noting that $$f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x-h)}{2h}\tag2$$ for all $x\in\mathbb R$ ...
Replacing $h$ with $-h$, we see that equation $(1)$ holds for any $h \in \Bbb R$ and any $x \in (x_0- \delta, x_0 + \delta)$.
Therefore for any $\epsilon \in (-\delta, \delta)$ and any $h \in \Bbb R$, we have $$f((x_0 + \epsilon) + h) = -f((x_0 + \epsilon) - h) = -f(x_0 + (\epsilon - h)) = f(x_0 - (\epsilon - h))$$ valid for all $h \in \Bbb R$. Putting $h = x - x_0 - \epsilon$ in the above equation gives $$f(x) = f(x - 2\epsilon)$$ valid for all $x \in \Bbb R$ and all $\epsilon \in (-\delta, \delta)$.
From here it should be clear that $f$ is constant and hence must be equal to $0$ by $(1)$.