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

Just as you replaced $x$ by $\frac{1}{x}$, you can replace $x$ by $1-x$. Then you can replace $x$ by $\frac{1}{x}$ in this new equation.
Repeating this process, I believe you will obtain 6 independent equations in the variables $f(z)$ where $z$ takes the values $$x,\frac{1}{x},\frac{1}{1-x},\frac{x}{x-1},\frac{x-1}{x},1-x.$$
These can be solved for $f(x)$.
Carrying out this process gives $$24f(x)=-3x+\frac{3}{x}+\frac{1}{1-x}+\frac{7x}{x-1}+\frac{x-1}{x}-5(1-x)$$$$f(x)=\frac{2x^3+x^2+5x-2}{24x(x-1)}$$
