Finding function $f(x)$ which satisfy given functional equation 
Find all function $f:\mathbb{R}-\{0,1\}$ in $$f(x)+2f\left(\frac{1}{x}\right)+3f\left(\frac{x}{x-1}\right)=x$$

Attempt: put $\displaystyle x = \frac{1}{x}$, then $$f\left(\frac{1}{x}\right)+2f(x)+3f\left(\frac{1}{1-x}\right) = \frac{1}{x}$$
could some help me how to solve it , thanks

Subtract twice from what you get from the given equation to get:
$$-f(x)  + f\left(\frac{x}{x-1}\right) - 2f\left(\frac{1}{1-x} \right) = \frac{x^2 - 2}{3x}$$
Then in this equation plug $x= \frac {x}{x-1}$ to get:
$$-f\left(\frac {x}{x-1}\right) + f\left(x \right) - 2f\left(1-x\right) = \frac{-x^2+4x-2}{3x(x-1)}$$
Adding the two equations you will get:
$$- 2f\left(\frac{1}{1-x} \right) - 2f(1-x) = \frac{x^3 - 2x^2 + 2x}{3x(x-1)}$$
Substituting $x = 1-x$ you should be able to get:
$$f(x) + f\left(\frac{1}{x} \right) = \frac{-x^3 + x^2 - x +1}{6x(1-x)}$$
Substitute in the main equation to get:
$$f\left(\frac{1}{x} \right) + 3f\left(\frac{x}{x-1}\right) = x + \frac{x^3 - x^2 + x -1}{6x(1-x)}$$
Substitute $x = \frac 1x$ to get:
$$f\left(x\right) + 3f\left(\frac{1}{1-x}\right) = \frac{-x^3 + x^2 + 5x -5}{6x(x-1)}$$

From now on I will stop writing the RHS, as the caclualtions get messy. We substitute $x = \frac{1}{1-x}$ and subtract three times the new equations from the one above the get:
$$f(x) - 9f\left(\frac{x-1}{x}\right) = \cdots$$
Substitute $x = \frac{x-1}{x}$ and add nine times the new equation to the one above to get:
$$f(x)- 81f\left(\frac{1}{1-x}\right) = \cdots$$
Now combining the last equation with the one just above the line I draw you can get an expression for $f(x)$.
