Find $f(x)$ if $f(x)+f\left(\frac{1-x}{x}\right)=1-x$ If $f: \mathbb{R} \to\mathbb{R} $, $x \ne 0,1$
Find all functions $f(x)$ such that $f(x)+f\left(\frac{1-x}{x}\right)=1-x$
My try:
Letting $$g(x)=x+f(x)$$ we get
$$g(x)+g\left(\frac{1-x}{x}\right)=\frac{1}{x}$$
Replacing $x \to 1-x$, we get:
$$g(1-x)+g\left(\frac{x}{1-x}\right)=\frac{1}{1-x}$$
any clue here?
 A: Chasing that idea in the comments: $f(x)+f\left(\frac{1-x}{x}\right) = 1-x$, then $f\left(\frac{1-x}{x}\right) + f\left(\frac{2x-1}{1-x}\right) = \frac{2x-1}{x}$, then  $f\left(\frac{2x-1}{1-x}\right) + f\left(\frac{2-3x}{2x-1}\right) = \frac{2-3x}{1-x}$ and so on. It doesn't loop back.
In fact, there are only two points for which the sequence we get from iterating $T(x) =\frac{1-x}{x}$ ever repeats at all; its two fixed points $\frac{-1\pm\sqrt{5}}{2}$. We have $T^n(x)=\frac{F_n-F_{n+1}x}{F_nx -F_{n-1}}$ where $F_n$ is the Fibonacci sequence; solving $T^n(x)=x$, we get the quadratic equation $F_nx^2 -F_{n-1}x = F_n-F_{n+1}x$, which simplifies to $F_n(x^2+x-1)=0$.
Because of this behavior, we can construct infinitely many very badly behaved functions that satisfy the functional equation. Choose some arbitrary $x$, choose $f(x)$ arbitrarily, define $f(Tx) = 1-x-f(x)$, define $f(T^{-1}x)=1-T^{-1}x-f(x)$, and keep iterating in both directions. Repeat this process on new values of $x$ until everything is filled in.
There's one place that demands special treatment - the sequence $$\dots,\frac58,\frac35,\frac23,\frac12,1,0,\infty,-1,-2,-\frac32,-\frac53,-\frac85,\dots$$
Here, we can't iterate in both directions, because of that $\infty$. Instead, choose an arbitrary value for $f(-1)$ and iterate forward only, and an arbitrary value for $f(1)$ and iterate backward only. We do need that value of $f(1)$ to make sense of the functional equation $f\left(\frac12\right)+f(1)=\frac12$ at $\frac12$; this problem doesn't lend itself to neatly cutting off the domain of $f$ around the bad values.
That key sequence can be used to cut $\mathbb{R}$ into regions that each only appear once in each sequence of iterates of $T$. Our choices can be condensed to a choice of an arbitrary function on $[1,\infty)$ and a choice of an arbitrary value at $-1$.
This problem simply doesn't have a nice answer. It's quite likely that the question is in error, and the transformation $T$ was meant to be something of finite order like $\frac{x-1}{x}$.
