How to solve $f'(x)=f(-x)$? By inspection,  $f'(x)=f(-x)$ seems only has trivial solution $f(x)=0$, is it right or wrong?
Let's assume in addition that $f$ is twice-differentiable. Then the solutions are precisely the functions of the form
$$
f(x) = A(\sin(x) + \cos(x)).
$$

Proof: taking another derivative, we see
$$
f''(x) = -f'(-x) = -f(x).
$$
This latter condition is satisfied precisely by functions of the form $A\sin(x) + B\cos(x)$, but most of these are not solutions to the original equation. Indeed, if $f(x) = A\sin(x) + B\cos(x)$, then $f(-x) = -A\sin(x) + B\cos(x)$ whereas $f'(x) = A\cos(x) - B\sin(x)$, so we get a solution if and only if $A = B$.

There may be exotic solutions that are once but not twice-differentiable; I'm not sure. (EDIT: there are no exotic solutions, see the comment.)
The general solution to this problem is $C(\sin x + \cos x)$, to prove it, just use $2f(x) = f(x)+f(-x)+f(x)-f(-x)$