A functional equation: $f\left(x^2-1\right)+2f\left(\frac{2x-1}{(x-1)^2}\right)=2-\frac{4}{x}+\frac{3}{x^2}$ 
If for all $ x > 1 $
$$
f\left(x^2-1\right)+2f\left(\frac{2x-1}{(x-1)^2}\right)=2-\frac{4}{x}+\frac{3}{x^2} \text,
$$
then $f(x)=?$

I don't know how to solve such equations. Help me please. Thank you.
 A: The solution is based on finding a change of variable $x\to\phi(x)$ such that $\phi(\phi(x))=x$ leading to a system of 2 equations in 2 variables.
Let's rewrite
$$\frac{2x-1}{\left(x-1\right)^{2}}=\frac{-x^{2}+2x-1+x^{2}}{\left(x-1\right)^{2}}=\frac{x^{2}-\left(x-1\right)^{2}}{\left(x-1\right)^{2}}=\frac{x^{2}}{\left(x-1\right)^{2}}-1$$
Which suggests that
$$x\to\frac{x}{x-1}$$
might be a good candidate. Indeed
$$\frac{2\left(\frac{x}{x-1}\right)-1}{\left(\frac{x}{x-1}-1\right)^{2}}=x^{2}-1$$
Making the change and not remembering to do this in the RHS as well we obtain:
$$f\left(\frac{2x-1}{\left(x-1\right)^{2}}\right)+2f(x^2-1)=2-\frac{4}{x}\left(x-1\right)+\frac{3}{x^{2}}\left(x-1\right)^{2}$$
Now multiply this equation by 2 and subtract the original one
$$3f(x^{2}-1)=2-\frac{4}{x}\left(2\left(x-1\right)-1\right)+\frac{3}{x^{2}}\left(2\left(x-1\right)^{2}-1\right)=\\2-\frac{4}{x}\left(2x-3\right)+\frac{3}{x^{2}}\left(2x^{2}-4x+1\right)=\\2-8+\frac{12}{x}+6-\frac{12}{x}+\frac{3}{x^{2}}=\frac{3}{x^{2}}$$
$$f(x^{2}-1)=\frac{1}{(x^{2}-1)+1}$$
$$f(x)=\frac{1}{x+1}$$
