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 $f(-1) \neq -\frac{1}{2}$, 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 $f(-1) = -\frac{1}{2}$.
$$x = 0 \Rightarrow f(f(y)) = (f(0) + \frac{1}{2})(f(y) + \frac{1}{2}). \quad (1)$$
Substituting $y = -1$ in (1) we see that $f(-\frac{1}{2}) = 0$. Now suppose that for some $y_0 \neq -1$ 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 $f(y) = -\frac{1}{2} \Leftrightarrow y = -1$. Now take some $y \neq -1$, and substitute in the given equality
$$x = \frac{-\frac{1}{2} - f(y)}{y + 1} \text{. Then } 0 = f(-\frac{1}{2}) = (f(y) + \frac{1}{2}) \left( f\left(\frac{-\frac{1}{2} - f(y)}{y+1}\right) + \frac{1}{2} \right) \text{. Since } y \neq -1, \\ f(y) + \frac{1}{2} \neq 0, \text{ and so } f\left(\frac{-\frac{1}{2} - f(y)}{y+1}\right) = -\frac{1}{2} \Rightarrow \frac{-\frac{1}{2} - f(y)}{y+1} = -1 \Rightarrow f(y) = y + \frac{1}{2} \text{. An easy}$$
check shows that this function meets all the requirements.