Substituting $x = 0$ into the equation gives $f(y) = f(y^2) + f(y)$, implying $f(y^2) = 0$. Thus $f(x) = 0$ for all non-negative real numbers $x$. In particular $f(x^2 + y^2) = 0$, which allows to simplify the initial equation as
$$f(xf(y) + y) = f(y).$$
Suppose that $f(c) \neq 0$ for some negative real number $c$. Substituting $y = c$ into (3) gives $f(z) = f(c)$ for all $z$ because the expression $xf(c) + c$ obtains all real values. By the above, there exists $z$ such that $f(z) = 0$; hence $f(c) = 0$, contradicting the choice of $c$. Consequently, $f(x) = 0$ for every real number $x$. Clearly the function $f(x) = 0$ satisfies the given equation.