$x\rightarrow \frac{x-3}{x+1}\Rightarrow f(\frac{3+x}{1-x})+f(x)=\frac{x-3}{x+1}$ $x\rightarrow \frac{x+3}{1-x}\Rightarrow f(\frac{x-3}{x+1})+f(x)=\frac{x+3}{1-x}$ $\Rightarrow f(x)=\frac{x^3+7x}{2(1-x^2)}$

See here: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=36&t=329994&p=1766069&hilit=Functional+Equation#p1766069

Here is my solution for this problem Solution $f\Big(\dfrac{x - 3}{x + 1}\Big) + f\Big(\dfrac{x + 3}{1 - x}\Big) = x$ Let $u = \dfrac{x - 3}{x + 1}$ then: $x = \dfrac{u + 3}{1 - u}$ and $\dfrac{x - 3}{x + 1} = \dfrac{u - 3}{u + 1}$ So: $f(u) + f\Big(\dfrac{u - 3}{u + 1}\Big) = \dfrac{u + 3}{1 - u}$ or $f(x) + f\Big(\dfrac{x - 3}{x + 1}\Big) = \dfrac{x + 3}{1 - x}, \forall x \not \in \{- 1; 1\}$ Similarly: Let $v = \dfrac{x + 3}{1 - x}$ then: $f(x) + f\Big(\dfrac{x + 3}{1 - x}\Big) = \dfrac{x - 3}{x + 1}, \forall x \not \in \{- 1; 1\}$ Hence: $2 f(x) = \dfrac{x + 3}{1 - x} + \dfrac{x - 3}{x + 1} - f\Big(\dfrac{x - 3}{x + 1}\Big) - f\Big(\dfrac{x + 3}{1 - x}\Big) = \dfrac{x + 3}{1 - x} + \dfrac{x - 3}{x + 1} - x = \dfrac{x^3 + 7x}{1 - x^2}$ or $f(x) = \dfrac{x^3 + 7x}{2 - 2x^2}, \forall x \not \in \{- 1; 1\}$ Retry, we see that: $f(x) = \dfrac{x^3 + 7x}{2 - 2x^2}$ satisfies the problem In conclusion, we have: $f(x) = \dfrac{x^3 + 7x}{2 - 2x^2}, \forall x \not \in \{- 1; 1\}$