$x \to 1 - \frac{1}{x}$ yields,
$f\left( {1 - \frac{1}{x}} \right) + f\left[ {\frac{1}{{1 - \left( {1 - \frac{1}{x}} \right)}}} \right] = 1 - \frac{1}{x}$
$ \Rightarrow f\left( {\frac{{x - 1}}{x}} \right) + f(x) = \frac{{x - 1}}{x}........(A)$
$x \to \frac{1}{{1 - x}}$ yields,
$f\left( {\frac{1}{{1 - x}}} \right) + f\left( {\frac{1}{{1 - \frac{1}{{1 - x}}}}} \right) = \frac{1}{{1 - x}}$
$ \Rightarrow f\left( {\frac{1}{{1 - x}}} \right) + f\left( {\frac{{x - 1}}{x}} \right) = \frac{1}{{1 - x}}........(B)$
(A) - (B) yields,
$f(x) - f\left( {\frac{1}{{1 - x}}} \right) = \frac{{x - 1}}{x} - \frac{1}{{1 - x}}........(C)$
(C) + The original equation yields,
$2f(x) = x + \frac{{x - 1}}{x} - \frac{1}{{1 - x}} = \frac{{{x^2}(1 - x) - {{(1 - x)}^2} - x}}{{x(1 - x)}}$
$ \Rightarrow f(x) = \frac{{{x^2} - {x^3} - 1 - {x^2} + 2x - x}}{{2x(1 - x)}} = \frac{{ - {x^3} + x - 1}}{{2x(1 - x)}}$