$x \to 1 – \frac{1}{x}$ 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)}}$