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?
 A: 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.)
A: As to the differentiability assumption, you can avoid assuming twice differentiability by considering $g(x)=f(-x)$. Then
$$
f'(x)=g(x)\\
g'(x)=-f(x)
$$
is a linear system of dimension $2$, which by the existence theorems has a global solution. As linear functions are infinitely smooth, a consequence of the existence of the solution is that the solutions are also differentiable of any order (analytical actually).
