<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] $f\left(\dfrac{x-3}{x+1}\right)+f\left(\dfrac{x+3}{x+1}\right)=x$
$f\left(1-\dfrac{4}{x+1}\right)+f\left(1+\dfrac{2}{x+1}\right)=x$
$f\left(1-\dfrac{4}{x}\right)+f\left(1+\dfrac{2}{x}\right)=x-1$
$f(1-2x)+f(1+x)=\dfrac{2}{x}-1$
$f(3-2x)+f(x)=\dfrac{2}{x-1}-1~......(1)$
$f(3-2(3-2x))+f(3-2x)=\dfrac{2}{3-2x-1}-1$
$f(4x-3)+f(3-2x)=-\dfrac{1}{x-1}-1~......(2)$
$(2)-(1)$ :
$f(4x-3)-f(x)=-\dfrac{3}{x-1}$
$f(4(4^x+1)-3)-f(4^x+1)=-\dfrac{3}{4^x+1-1}$
$f(4^{x+1}+1)-f(4^x+1)=-3\times4^{-x}$
$f(4^x+1)=\Theta(x)+4^{1-x}$ , where $\Theta(x)$ is an arbitrary periodic function with unit period
$f(x)=\Theta(\log_4(x-1))+\dfrac{4}{x-1}$ , where $\Theta(x)$ is an arbitrary periodic function with unit period
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