$f(x)-1=f(x-1)=(x-1)^2f\left(\frac{1}{x-1}\right)=(x-1)^2\left(f\left(\frac{x}{x-1}\right)-1\right)=(x-1)^2\cdot\frac{x^2}{(x-1)^2}f\left(\frac{x-1}{x}\right)-(x-1)^2=x^2f\left(1-\frac{1}{x}\right)-(x-1)^2=x^2-x^2f\left(\frac{1}{x}\right)-(x-1)^2=2x-1-f(x)\implies f(x)=x$