For $y=0$ from the condition we have, that $f(f(x)+f(0))=0$ for any $x$, and there is such $a$ that $f(a)=0$. Thus, for $x=a$, $f(f(0))=0$. Then, for $x=y=0$ it follows, that $f(2f(0))=0$, and for $x=y=f(0)$ we have, that
$$f(f(f(0)) + f(f(0))) = f^2(0)f(2f(0)) = 0,$$
i.e. $f(0)=0$ and $f(f(x))=0$ for any $x$.

Let $y=f(y)$ in the condition, then
$$xf(y)f(x+f(y)) = f(f(x)+f(f(y))) = f(f(x)) = 0,$$
so $xf(y)f(x+f(y))=0$ for arbitrary $x$ and $y$.

Let $f(y_0) \neq 0$ for some $y_0 > 0$. Let $y = y_0 - x$, then
$$f(f(x) + f(y_0 - x)) = x(y_0 - x)f(y_0).$$
Expression in right part takes arbitrary value in segment $[0; \frac{1}{4}y_0^2 f(y_0)]$, so range of function $f$ includes this segment. Thus, there is some $x_0$ such, that $f(x_0) = \min\{\frac{1}{2}y_0, \frac{1}{8}y_0^2 f(y_0)\}$. From equality $xf(y)f(x+f(y))=0$ for $y=x_0$ and $x=y_0-f(x_0)>0$ we have, that $f(x_0)f(y_0)=0$, that is impossible. This contradiction proves, that $f(x)=0$ for arbitrary $x \in [0, +\infty)$.