Let : $$f(x)+f\left(\frac{1-x}{x} \right)=x$$ then: $$f(x)=?$$
My Try :
$x\to \dfrac{1-x}{x}$
\begin{align}f\left(\dfrac{1-x}{x}\right)+f\left(\frac{1-\dfrac{1-x}{x}}{\dfrac{1-x}{x}} \right)&=\dfrac{1-x}{x}\\ f\left(\dfrac{1-x}{x}\right)+f\left(\dfrac{2x-1}{x^2} \right)&=\dfrac{1-x}{x}\end{align}
And $u=\dfrac{1-x}{x}$
$$f(u)+f\left(\dfrac{2x-1}{x^2} \right)=u$$
Now what ?