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.