Find function $ f $ such that $f(\frac{x-3}{x+1})+f(\frac{x+3}{1-x})=x$ I am looking for functions $ f:\Bbb R \to \Bbb R $ satisfying
$$f\Big(\frac{x-3}{x+1}\Big)+f\Big(\frac{x+3}{1-x}\Big)=x$$
I used the substitution $ x=\cos(2t) $ for $ x\in (0,2\pi) $, to use the identities
$$1+x=2\cos^2(t) \text{ and } 1-x=2\sin^2(t)$$
The new equation will be
$$f\Big(1-\frac{2}{\cos^2(t)}\Big)+f\Big(\frac{2}{\sin^2(t)}-1\Big)=$$
$$\cos^2(t)-\sin^2(t)$$
I think that this approach won't allow me the get the answer.
Any idea will be appreciated.
 A: Substitute in $x\mapsto\frac{x-3}{x+1}$. Then we have
$$f\left(\frac{x+3}{1-x}\right)+f\left(x\right)=\frac{x-3}{x+1}$$
Now substitute again $x\mapsto \frac{x-3}{x+1}$,
$$f(x)+f\left(\frac{x-3}{x+1}\right)=\frac{x+3}{1-x}$$
Now add these two to get
$$2f(x)+f\left(\frac{x+3}{1-x}\right)+f\left(\frac{x-3}{x+1}\right)=\frac{x-3}{x+1}+\frac{x+3}{1-x}$$
$$2f(x)=\frac{x-3}{x+1}+\frac{x+3}{1-x}-x$$
$$f(x)=\frac{1}{2}\left(\frac{x-3}{x+1}+\frac{x+3}{1-x}-x\right)$$
A: Let $h(x)=f(\frac{x-3}{x+1})$. Then the functional equation becomes
$$\tag1 h(x)+h(-x)=x\qquad\text{for }x\ne\pm1.$$
There are obviously many such $h$. In fact, if $k\colon(0,\infty)\setminus\{1\}\to\Bbb R$ is arbitrary, we can set
$$h(x)=\begin{cases}k(x)&x>0\\x-k(x)&x<0\\0&x=0\end{cases} $$
and obtain a solution for $(1)$.
As the inverse of $x\mapsto \frac{x-3}{x+1}$ is (also) $x\mapsto \frac{x+3}{1-x}$, we obtain a solution
$$ f(x)=h(\tfrac{x+3}{1-x})$$
of the original functional equation.
A: Set $u= \frac{x-3}{x+1}$ the equation becomes $f(u)+f(\frac{u+3}{1-u})=\frac{u+3}{1-u}$
Set $v=\frac{x+3}{1-x}$ the equation becomes $f(\frac{v-3}{v+1})+f(v)=\frac{v-3}{v+1}$
Now replace all dummy variables with $y$ and solve for $f(x)$,

*

*$$A+B=y$$
2.$$B+C=\frac{y+3}{1-y}$$


*$$A+C=\frac{y-3}{y+1}$$
Can you finish it?

