<s>[INST] 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.
 [/INST] 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
$$2f(x)+f\left(\frac{x+3}{1-x}\right)+f\left(\frac{x-3}{x+1}\right)=\frac{x-3}{x+1}+\frac{x+3}{1-x}$$
$$2f(x)=\frac{x-3}{x+1}+\frac{x+3}{1-x}-x$$
$$f(x)=\frac{1}{2}\left(\frac{x-3}{x+1}+\frac{x+3}{1-x}-x\right)$$
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