<s>[INST] Finding $f(x)$ from a equation $f(x)+f(\frac{x-1}{x}) = x+1$ is given.  
find $f(x)$. 
I did tried to change it into another form buy substituting $x$ with $\frac{x}{x-1}$  
The result were $f(\frac{1}{x})+f(\frac{x}{x-1})=\frac{2x-1}{x-1}$  
I don't know what to do next.
 [/INST] All you have to do is make the substitution again.
$$f(x)+f\bigg(\frac{x-1}{x}\bigg)=x+1$$
$$\implies f\bigg(\frac{x-1}{x}\bigg)+f\bigg(\frac{1}{1-x}\bigg)=\frac{x-1}{x}+1$$
$$\implies f\bigg(\frac{1}{1-x}\bigg)+f(x)=\frac{1}{1-x}+1$$
Now subtract the first and second equations to get
$$f(x)-f\bigg(\frac{1}{1-x}\bigg)=x-\frac{x-1}{x}$$
Then add this to the third equation to get
$$f(x)+f(x)=x-\frac{x-1}{x}+\frac{1}{1-x}+1$$
$$2f(x)=x-\frac{x-1}{x}+\frac{1}{1-x}+1$$
$$2f(x)=x-1+\frac{1}{x}+\frac{1}{1-x}+1$$
$$2f(x)=x+\frac{1}{x}+\frac{1}{1-x}$$
$$2f(x)=x+\frac{1}{x(1-x)}$$
$$\color{red}{f(x)=\frac{x^2(1-x)+1}{2x(1-x)}}$$
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