49th Mathematical Olympiad in Ukraine

If we take $y=-1$ we'll get that: $f(f(-1))=(f(x)+\frac{1}{2})(f(-1)+\frac{1}{2})$. So if $f(-1)\ne -\frac{1}{2}$, then $f$ is constant. If we substitute $f=c$ in our equality, then we'll get that $c=(c+\frac{1}{2}{)}^2$ which is impossible. Therefore $f(-1)=-\frac{1}{2}$. $$x=0\Rightarrow f(f(y))=(f(0)+\frac{1}{2})(f(y)+\frac{1}{2}).(1)$$ Substituting $y=-1$ in (1) we see that $f(-\frac{1}{2})=0$. Now suppose that for some $y_0\ne -1$ we have $f(y_0)=-\frac{1}{2}$. Then substituting $y=y_0$ in the given equality we will obtain: $f(x(1+y_0)-\frac{1}{2})=0$, which means that $f$ is constant which was proved to be impossible. Thus $f(y)=-\frac{1}{2}\iff y=-1$. Now take some $y\ne -1$, and substitute in the given equality $$\begin{gathered} x=\frac{-\frac{1}{2}-f(y)}{y+1}\text{. Then }0=f(-\frac{1}{2})=(f(y)+\frac{1}{2})\left(f\left(\frac{-\frac{1}{2}-f(y)}{y+1}\right)+\frac{1}{2}\right)\text{. Since }y\ne -1, \\ f(y)+\frac{1}{2}\ne 0,\text{ and so }f\left(\frac{-\frac{1}{2}-f(y)}{y+1}\right)=-\frac{1}{2}\Rightarrow \frac{-\frac{1}{2}-f(y)}{y+1}=-1\Rightarrow f(y)=y+\frac{1}{2}\text{. An easy} \end{gathered}$$ check shows that this function meets all the requirements.