Let $\mathbb{R}^*$ be the set of all real numbers, except $1$. Find all functions $f:\mathbb{R}^* \rightarrow \mathbb{R}$ that satisfy the functional equation $$x+f(x)+2f\left(\frac{x+2009}{x-1}\right)=2010$$.
1. Given the functional equation:
   \[
   x + f(x) + 2f\left(\frac{x+2009}{x-1}\right) = 2010
   \]
   for \( f: \mathbb{R}^* \rightarrow \mathbb{R} \), where \( \mathbb{R}^* = \mathbb{R} \setminus \{1\} \).

2. To find \( f(x) \), we substitute \( x \) with \( \frac{x+2009}{x-1} \) in the original equation. Let \( y = \frac{x+2009}{x-1} \). Then:
   \[
   y + f(y) + 2f(x) = 2010
   \]
   This gives us two equations:
   \[
   x + f(x) + 2f\left(\frac{x+2009}{x-1}\right) = 2010 \quad \text{(1)}
   \]
   \[
   \frac{x+2009}{x-1} + f\left(\frac{x+2009}{x-1}\right) + 2f(x) = 2010 \quad \text{(2)}
   \]

3. From equation (1), we have:
   \[
   x + f(x) + 2f\left(\frac{x+2009}{x-1}\right) = 2010
   \]
   From equation (2), we have:
   \[
   \frac{x+2009}{x-1} + f\left(\frac{x+2009}{x-1}\right) + 2f(x) = 2010
   \]

4. Subtract equation (2) from equation (1):
   \[
   \left( x + f(x) + 2f\left(\frac{x+2009}{x-1}\right) \right) - \left( \frac{x+2009}{x-1} + f\left(\frac{x+2009}{x-1}\right) + 2f(x) \right) = 0
   \]
   Simplifying, we get:
   \[
   x + f(x) + 2f\left(\frac{x+2009}{x-1}\right) - \frac{x+2009}{x-1} - f\left(\frac{x+2009}{x-1}\right) - 2f(x) = 0
   \]
   \[
   x - \frac{x+2009}{x-1} + f(x) - 2f(x) + 2f\left(\frac{x+2009}{x-1}\right) - f\left(\frac{x+2009}{x-1}\right) = 0
   \]
   \[
   x - \frac{x+2009}{x-1} - f(x) + f\left(\frac{x+2009}{x-1}\right) = 0
   \]

5. Rearrange to solve for \( f(x) \):
   \[
   f(x) = x - \frac{x+2009}{x-1} + f\left(\frac{x+2009}{x-1}\right)
   \]

6. Let \( z = \frac{x+2009}{x-1} \). Then:
   \[
   f(x) = x - z + f(z)
   \]
   Substitute \( z \) back:
   \[
   f(x) = x - \frac{x+2009}{x-1} + f\left(\frac{x+2009}{x-1}\right)
   \]

7. To find \( f(x) \), we need to simplify the expression:
   \[
   f(x) = x - \frac{x+2009}{x-1} + f\left(\frac{x+2009}{x-1}\right)
   \]
   \[
   f(x) = x - \frac{x+2009}{x-1} + f\left(\frac{x+2009}{x-1}\right)
   \]
   \[
   f(x) = x - \frac{x+2009}{x-1} + f\left(\frac{x+2009}{x-1}\right)
   \]

8. Simplify \( \frac{x+2009}{x-1} \):
   \[
   \frac{x+2009}{x-1} = \frac{x-1+2010}{x-1} = 1 + \frac{2010}{x-1}
   \]

9. Substitute back:
   \[
   f(x) = x - \left(1 + \frac{2010}{x-1}\right) + f\left(1 + \frac{2010}{x-1}\right)
   \]
   \[
   f(x) = x - 1 - \frac{2010}{x-1} + f\left(1 + \frac{2010}{x-1}\right)
   \]

10. To find a specific form of \( f(x) \), assume \( f(x) = \frac{1}{3}(x + 2010 - 2\frac{x+2009}{x-1}) \):
    \[
    f(x) = \frac{1}{3}\left(x + 2010 - 2\left(1 + \frac{2010}{x-1}\right)\right)
    \]
    \[
    f(x) = \frac{1}{3}\left(x + 2010 - 2 - \frac{4020}{x-1}\right)
    \]
    \[
    f(x) = \frac{1}{3}\left(x + 2008 - \frac{4020}{x-1}\right)
    \]

11. Verify the solution by substituting back into the original equation:
    \[
    x + \frac{1}{3}\left(x + 2008 - \frac{4020}{x-1}\right) + 2\left(\frac{1}{3}\left(\frac{x+2009}{x-1} + 2008 - \frac{4020}{\frac{x+2009}{x-1}-1}\right)\right) = 2010
    \]

12. Simplify and verify that the equation holds true.

The final answer is \( \boxed{ f(x) = \frac{1}{3}\left(x + 2010 - 2\frac{x+2009}{x-1}\right) } \).