Ukrainian National Mathematical Olympiad, 3rd Round

Take $x = y$, we get:
$$f(2x) = f(0)f(2x) + f(0)f(-2x). \quad (1)$$
Take in (1) $x = 0$ and we get $f(0) = 2f^2(0)$. So $f(0) = 0$ or $f(0) = \frac{1}{2}$.
If $f(0) = 0$, (1) implies that $f(2x) = 0$, hence $f = 0$.

If $f(0) = \frac{1}{2}$, (1) implies that $f(2x) = \frac{1}{2}f(2x) + \frac{1}{2}f(-2x)$ and $f(2x) = f(-2x)$.
Taking $x = 0$ and using the fact that our function is even, we arrive at
$$\frac{1}{2} = f(0) = (f(y))^2 + (f(-y))^2 = 2(f(y))^2.$$
This gives us that $f = \frac{1}{2}$.

It is easy to check that both functions satisfy all the requirements.