<s>[INST] Functions satisfying $f(x)+f(\frac{1}{1-x})=x$ with $x\in\mathbb{R}\setminus\{0,1\}.$ I have used this identity: if $g(x)=1/(1-x),$ then
$$g^{-1}(x)=1-\frac{1}{x},$$ 
to get all functions satisfying: $f(x)+f(\frac{1}{1-x})=x$ with $x\in\mathbb{R}\setminus\{0,1\},$ but I didn't get a general form of its solution. My question here is:

Is there any simple method to solve the titled functional equation? 

 [/INST] We will prove that $$f(x)=x-\frac{1}{2(1-x)}$$ when
$$f(x)+f\left(\frac{1}{1-x}\right)=x$$
At first we get $$f\left(\frac{1}{1-x}\right)+f\left(\frac{1}{1-\frac{1}{1-x}}\right)=\frac{1}{1-x}$$ and we get
$$f\left(\frac{1}{1-x}\right)+f\left(\frac{x-1}{x}\right)=\frac{1}{1-x}$$
Now we get
$$f\left(\frac{x-1}{x}\right)+f\left(\frac{1}{1-\frac{x-1}{x}}\right)=\frac{x-1}{x}$$
so
$$f\left(\frac{x-1}{x}\right)+f(x)=\frac{x-1}{x}$$
and we obtain
using that
$$f\left(\frac{1}{1-x}\right)=x-f(x)$$
$$x-f(x)+\frac{x-1}{x}-f(x)=\frac{1}{1-x}$$
so $$2f(x)=x+\frac{x-1}{x}-\frac{1}{1-x}$$
and $$f(x)=\frac{-x^3+x-1}{x}$$
</s>