Prove there are no other solutions of functional equation $f(x+y) = \frac{f(x)+f(y)}{1-f(x)f(y)}$ I have the following functional equation. Find all continuous functions $f:(-1,1) \to \mathbb R$ such that
$$
f(x+y)=\frac{f(x) + f(y)}{1 - f(x)f(y)}
$$
The first obvious solution is $f(x) \equiv 0$. Another one I guessed, it is $f(x) = \pm \tan x$. I suspect, that there are no more solutions. The problem is that I don't know how to prove that.
Since we have a rational equation, I have no idea how to make any substitutions in order to get expression for $\tan x$ (as it is not a rational expression).
P.S. I can also show that it is true that (a) $f(0) = 0$ (take $x=y=0$), and (b) $f(-x) = -f(x)$ (take $y=-x$).
Update: there is a solution of this problem also here: https://artofproblemsolving.com/community/c6h386060
 A: Define $g(x)=\arctan(f(x))$, so that $f(x)=\tan(g(x))$. Then $g:(-1,1)\to(-\pi/2,\pi/2)$ is continuous.  The functional equation becomes
$$
\tan(g(x+y))=\frac{\tan(g(x))+\tan(g(y))}{1-\tan(g(x))\tan(g(y))}=\tan(g(x)+g(y)).
$$
The latter equality used the tangent angle addition formula. It follows that $g(x+y)-g(x)-g(y)\in \mathbb{Z}\cdot\pi$. The function $w(x,y):=g(x+y)-g(x)-g(y)$ is continuous and discretely-valued on the connected set $\{(x,y)\in(-1,1)^2:x+y\in(-1,1)\}$, so $w$ must be constant. Since $w(0,0)=0$, we have $w\equiv 0$, and we conclude that
$$
g(x+y)=g(x)+g(y),
$$
for all $x$, $y\in(-1,1)$ such that $x+y\in(-1,1)$. It is well-known that every continuous real-valued functions on $\mathbb{R}$ that preserves addition is of the form multiplication by a constant, and essentially the same proof works for functions on $(-1,1)$. I won't write out the details.
Since $g$ takes $(-1,1)$ into $(-\pi/2,\pi/2)$, we must have $g(x)=Cx$ for some $C\in [-\pi/2,\pi/2]$. We conclude that the only solutions to the original equation are
$$
f(x)=\tan(Cx)
$$
for $C\in[-\pi/2,\pi/2]$. The cases $C=0,\pm1$ are the solutions you found.
