<s>[INST] Solve $f\left(\frac{x-3}{x+1}\right)+f\left(\frac{x+3}{x+1}\right)=x$ $\forall x\neq -1$ Given function $y=f(x)$
such that 
$$f\left(\frac{x-3}{x+1}\right)+f\left(\frac{x+3}{x+1}\right)=x \quad\forall x\neq -1$$
find $f(x)$ and $f(2007)$.
 [/INST] With 
$$f(x) = \frac{x+3}{x-1}-\frac{3}{2}$$
we have
$$a := f\left(\frac{x-3}{x+1}\right) = -x-\frac{3}{2}$$
and 
$$b := f\left(\frac{x+3}{x+1}\right)=2x+\frac{3}{2}$$
Therefore the sum is $a+b=x$ and $f(2007)=-\frac{999}{2006}.$
Edit regarding the comments: My function $f(x)\,$ is a solution of the functional equation (FE). The road to it was as follows: If the FE is valid for all $x\ne -1\;$ and you plug in large $x$ into the FE, you see that there should be a singularity at $x=1.\;$ Making the ansatz $f(x) = \frac{g(x)}{x-1}+\cdots$ and playing arround with Maple arrives at the given function.
If the OP has a different function solving the FE, this only shows that there are at least two solutions.
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