, 2008

If $x = 0$ we get $f(0) = f(y) + y f(0)$. So, $f$ is a linear function of the form $f(x) = a - a x$ for some real $a$. Inserting this into the functional equation we see that for all $x, y \in \mathbb{R}$ we have $x + a - a x(a - a y) = a - a y + y(a - a x)$, so
$$x - a^2 x + a^2 x y = -a x y.$$
Now, let $y = 0$ to see that $x = a^2 x$ for all $x$, so $a^2 = 1$. Under this condition the above equality becomes $x y = -a x y$ or, equivalently, $(1 + a) x y = 0$, which then implies $a = -1$, since this last equality has to hold for all $x$ and $y$.
We have shown that $f(x) = x - 1$. Using this expression for $f$ in the initial functional equation, we see that the left-hand side is equal to
$$x + f(x f(y)) = x + f(x y - x) = x + x y - x - 1 = x y - 1$$
and the right-hand side becomes
$$f(y) + y f(x) = y - 1 + y(x - 1) = x y - 1.$$
The two sides are equal for all $x$ and $y$, so $f(x) = x - 1$ is the (only) solution to our equation.