Here is my attempt. Start by taking the derivative of both sides with respect to $x$:
$$\frac{d}{dx} f'(x) = \frac{d}{dx} f(-x).$$
This simplifies to:
$$f''(x) = -f'(-x).$$
Using the given equation we get
$$f''(x) = -f(x).$$
The general solution of this ode is
$$f(x)=A\cos x+ B\sin x,$$
where $(A,B)\in \mathbb{R}^2$.
Taking the first derivative results in
$$f'(x)=-A\sin x+ B\cos x.$$
For this to be equal to $f(-x)$ it suffices that $A=B$.
Hence; the set of functions satisfying the original equation reads finally
$$f(x)=A(\cos x +\sin x).$$
Questions: Is my solution rigorous? Other ways to find the requested set of functions?
To me your work looks correct, except at the end there's a sketchy part where you say "it suffices that $A=B$". That's true, but you also want it to be necessary, not just suffice, to claim that all solutions are of the form that you get. This can easily be verified by understanding that $f'(x) = f(-x)$ for all $x$, so in particular $f'(0) = f(0)$.