Considering $x = -f(y)$ we get $f(0) = f(f(y)) - (f(y))^2$. Denoting $f(0)$ by $a$ and substituting $y = 0$ we arrive at $f(a) = a^2 + a$. Now we return to the original equality and plug in $y = 0$:
$$f(x + a) = f(a) + 2x a + x^2 = x^2 + 2x a + a^2 + a = (x + a)^2 + a.$$
Thus, all the possible solutions are of the form $f(x) = x^2 + a$, where $a$ is a real constant. A routine check shows that every function of this form is indeed a solution to the original equation.