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.

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?

