Find all continuous functions $ f(x+y)=\frac{f(x)+f(y)+2 f(x) f(y)}{1-f(x) f(y)} $ question - 
Find all continuous functions $f: \mathbb{R} \rightarrow \mathbb{R}$ which satisfy the equation
$$
f(x+y)=\frac{f(x)+f(y)+2 f(x) f(y)}{1-f(x) f(y)}
$$
for all $x, y$
my try -
i proved that f(0)=0 ..then hint says substitute $g(x)=f(x) /(1+f(x))$..
i am not getting how to use this hint...i substitute in original equation and simplified lots of thing but nothing seems useful..
any hints ??? ...and how one can thought that we have to substitute this...
 A: Using the given hint, we get,
$$g(x)=\dfrac{f(x)}{1+f(x)}\\
\implies f(x)=\dfrac{g(x)}{1-g(x)}\\
\implies \dfrac{g(x+y)}{1-g(x+y)}=\dfrac{\dfrac{g(x)}{1-g(x)}+\dfrac{g(y)}{1-g(y)}+\dfrac{2g(x)g(y)}{(1-g(x))(1-g(y))}}{1-\dfrac{g(x)g(y)}{(1-g(x))(1-g(y))}}\\
\implies \dfrac{g(x+y)}{1-g(x+y)}=\dfrac{g(x)(1-g(y))+g(y)(1-g(x))+2g(x)g(y)}{1-g(x)-g(y)}\\
\implies\dfrac{g(x+y)}{1-g(x+y)}=\dfrac{g(x)+g(y)}{1-g(x)-g(y)}\\
\implies \dfrac{g(x+y)}{1-g(x+y)}+1=\dfrac{g(x)+g(y)}{1-g(x)-g(y)}+1\\
\implies \dfrac{1}{1-g(x+y)}=\dfrac{1}{1-g(x)-g(y)}\\
\implies \boxed{g(x+y)=g(x)+g(y)}$$
Now use this.
A: As suggested, let $g(x)=\frac{f(x)}{1+f(x)}$. Then $f(x)=\frac{g(x)}{1-g(x)}$. Thus
$$\frac{g(x+y)}{1-g(x+y)}=\frac{\frac{g(x)}{1-g(x)}+\frac{g(y)}{1-g(y)}+2\frac{g(x)}{1-g(x)} \cdot \frac{g(y)}{1-g(y)}}{1-\frac{g(x)}{1-g(x)} \cdot\frac{g(y)}{1-g(y)}}=\frac{g(x)+g(y)}{1-(g(x)+g(y))}.$$
So 
$$\frac{g(x+y)}{1-g(x+y)}=\frac{g(x)+g(y)}{1-(g(x)+g(y))}.$$
Since the function $\frac{t}{1-t}$ is monotone, we conclude from the above equality that 
$$g(x+y)=g(x)+g(y).$$
Thus $g(x)$ is a linear function with $g(0)=0$, and therefore $g(x)=mx$ for some $m \in \mathbb{R}$. Therefore if 
$$f(x)=\frac{mx}{1-mx}, m \in \mathbb{R},$$
then the identity in the statement of the problem will hold for points in the domain of the functions in the left and right hand sides, i.e. 
$$x,y \neq \frac{1}{m}, \ \ x+y \neq \frac{1}{m}.$$
If the problem really wants the identity to be satisfied for all $x,y \in \mathbb{R}$, then the domain of $f$ has to be all of $\mathbb{R}$, and hence the only function satisfying the assumptions is $f=0$ (m=0). 
