If we assume that $ \frac{f(y)-1}{y-1}>0$ so we can substitute $ x=\frac{f(y)-1}{y-1}$ and in this case $ y=\frac{f(x+f(y))}{f(xy+1)}=\frac{f(\frac{yf(y)-1}{y-1})}{f(\frac{yf(y)-1}{y-1})}=1$. So $ \frac{f(y)-1}{y-1}<0$ for every $ y\not=1$. (*) Now let $ x=1-\frac{1}{y}$ for $ y>1$. $ \Rightarrow f(1-\frac{1}{y}+f(y))=yf(y)$. If $ f(y)>\frac{1}{y}$, then $ 1+f(y)-\frac{1}{y}>1\Rightarrow f(1+f(y)-\frac{1}{y})<1$ from (*) - contradiction. If $ f(y)<\frac{1}{y}$, then $ 1+f(y)-\frac{1}{y}<1\Rightarrow f(1+f(y)-\frac{1}{y})>1$ from (*) - contradiction. $ \Rightarrow f(y)=\frac{1}{y}$ for every $ y>1$. Now, if we take some $ y>1$ and $ x>0$, we will have $ f(x+f(y))=f(x+\frac{1}{y})=yf(xy+1)=\frac{1}{x+\frac{1}{y}}$. So, if for some $ 0<a<b<1$ we substitute $ y=\frac{1}{b-a}>1$ and $ x=a$, then we will have $ f(b)=\frac{1}{b}$. $ \Rightarrow f(x)=\frac{1}{x}$ for every $ x>0$.

Can someone else post another solution? BG Yoda just adjusts the right answer. I want some more elegant method, if it's possible?

far-reaching wrote: Can someone else post another solution? BG Yoda just adjusts the right answer. I want some more elegant method, if it's possible? This is basically the same solution, written in different way: Take any $x>1$, we have \[f\left( \frac{x-1}{x}+f\left( x\right) \right) =xf\left( x\right) \] and \[f\left( x+f\left( \frac{x-1}{x}+f\left( x\right) \right) \right) =\left[ \frac{x-1}{x}+f\left( x\right) \right] f\left( x+xf\left( x\right) \right). \] Thus $\frac{x-1}{x}+f\left( x\right) =1$ or equivalently $f\left(x\right) =\frac{1}{x}$. Now for any $x>0$, we have \[\frac{1}{1+f\left( x\right) }=f\left( 1+f\left( x\right) \right) =xf\left(x+1\right) =\frac{x}{x+1}=\frac{1}{1+\frac{1}{x}}\] or $f\left( x\right) =\frac{1}{x}$.