24. (HUN 2) Determine all real functions $f(x)$ that are defined and continuous on the interval $(-1,1)$ and that satisfy the functional equation
$$
f(x+y)=\frac{f(x)+f(y)}{1-f(x) f(y)} \quad(x, y, x+y \in(-1,1)) .
$$
24. Setting $x=y=0$ gives us $f(0)=0$. Let us put $g(x)=\arctan f(x)$. The given functional equation becomes $\tan g(x+y)=\tan (g(x)+g(y))$; hence
$$
g(x+y)=g(x)+g(y)+k(x, y) \pi,
$$
where $k(x, y)$ is an integer function. But $k(x, y)$ is continuous and $k(0,0)=$ 0 , therefore $k(x, y)=0$. Thus we obtain the classical Cauchy's functional equation $g(x+y)=g(x)+g(y)$ on the interval $(-1,1)$, all of whose continuous solutions are of the form $g(x)=a x$ for some real $a$. Moreover, $g(x) \in(-\pi, \pi)$ implies $|a| \leq \pi / 2$.
Therefore $f(x)=\tan a x$ for some $|a| \leq \pi / 2$, and this is indeed a solution to the given equation.