Find the function $\displaystyle{f : \mathbb{R}-\left\{ 0\,\right\} \rightarrow \mathbb{R} }$ such that $$f(x)+f(1-\frac{1}{x}) = \frac{1}{x},\,\,\, \forall x \in \mathbb{R}- \{ 0, 1\,\}$$
[i](-InnoXenT-)[/i]
1. Given the functional equation:
   \[
   f(x) + f\left(1 - \frac{1}{x}\right) = \frac{1}{x}, \quad \forall x \in \mathbb{R} - \{0, 1\}
   \]
   we need to find the function \( f \).

2. Let's denote \( y = 1 - \frac{1}{x} \). Then:
   \[
   y = 1 - \frac{1}{x} \implies x = \frac{1}{1 - y}
   \]
   Substituting \( y \) back into the original equation, we get:
   \[
   f\left(\frac{1}{1 - y}\right) + f(y) = 1 - y
   \]

3. Now, consider the original equation and the transformed equation:
   \[
   f(x) + f\left(1 - \frac{1}{x}\right) = \frac{1}{x}
   \]
   \[
   f\left(1 - \frac{1}{x}\right) + f\left(\frac{x - 1}{x}\right) = 1 - x
   \]
   \[
   f\left(\frac{x - 1}{x}\right) + f\left(\frac{1}{1 - x}\right) = \frac{x}{x - 1}
   \]

4. Adding the first and third equations and subtracting the second equation, we get:
   \[
   \left(f(x) + f\left(1 - \frac{1}{x}\right)\right) + \left(f\left(\frac{x - 1}{x}\right) + f\left(\frac{1}{1 - x}\right)\right) - \left(f\left(1 - \frac{1}{x}\right) + f\left(\frac{x - 1}{x}\right)\right) = \frac{1}{x} + \frac{x}{x - 1} - (1 - x)
   \]

5. Simplifying the left-hand side:
   \[
   f(x) + f\left(\frac{1}{1 - x}\right) = \frac{1}{x} + \frac{x}{x - 1} - (1 - x)
   \]

6. Simplifying the right-hand side:
   \[
   \frac{1}{x} + \frac{x}{x - 1} - (1 - x) = \frac{1}{x} + \frac{x}{x - 1} - 1 + x
   \]
   \[
   = \frac{1}{x} + \frac{x}{x - 1} - 1 + x
   \]
   \[
   = \frac{1}{x} + \frac{x}{x - 1} - 1 + x
   \]

7. Combining the terms:
   \[
   = \frac{1}{x} + \frac{x}{x - 1} - 1 + x
   \]
   \[
   = \frac{1}{x} + \frac{x}{x - 1} - 1 + x
   \]

8. Therefore, we have:
   \[
   2f(x) = \frac{1}{x} + x - 1 + \frac{x}{x - 1}
   \]

9. Solving for \( f(x) \):
   \[
   f(x) = \frac{1}{2} \left( \frac{1}{x} + x - 1 + \frac{x}{x - 1} \right)
   \]

10. This function indeed fits the original equation for all \( x \neq 1 \). For \( x = 1 \), \( f(1) \) can be any real number \( a \).

The final answer is \( \boxed{ f(x) = \frac{1}{2} \left( \frac{1}{x} + x - 1 + \frac{x}{x - 1} \right) } \) for \( x \neq 1 \) and \( f(1) = a \) where \( a \in \mathbb{R} \).