Slovenija 2008

If $x=0$ we get $f(0)=f(y)+yf(0)$. So, $f$ is a linear function of the form $f(x)=a-ax$ for some real $a$. Inserting this into the functional equation we see that for all $x,y\in \mathbb{R}$ we have $x+a-ax(a-ay)=a-ay+y(a-ax)$, so $$x-a^{2}x+a^{2}xy=-axy.$$ 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 $xy=-axy$ or, equivalently, $(1+a)xy=0$, 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 $$x+f(xf(y))=x+f(xy-x)=x+xy-x-1=xy-1$$ and the right-hand side becomes $$f(y)+yf(x)=y-1+y(x-1)=xy-1.$$ The two sides are equal for all $x$ and $y$, so $f(x)=x-1$ is the (only) solution to our equation.