I am looking for all functions $f:(0,\infty)\to (0,\infty)$ that satisfy for all $x,y>0$: $$f(xy+f(x))=f(f(x)f(y))+x.$$
I found that $f(x)=x$ works. Are there other solutions?
I try setting $x=1$: $$f(y+f(1))=f(f(1)f(y))+1$$ and swapping $x,y$ to get $$f(xy+f(x))-f(xy+f(y))=x-y.$$ Now I don't know what to do.
$f(xy+f(x)) =f(f(x)f(y)) + x$
$f(xy+f(y))=f(f(x)f(y)) + y$
Subtracting with $y=0$ $f(f(x))=x+f(f(0))$
Now $f$ is bijective. Plugging $y=0$ again $f(f(x)) =x+f(f(x)f(0))$
Hence $f(0)=0$ because $f$ is bijective and $f(f(0))=f(f(x)f(0))$
$f(f(x)) =x$
Put $x=c$ such that $f(c) =1$, and $y:=y-c$ $f(cy+1-c^2)=y=f(f(y))$
$cy+1-c^2=f(y)$
$f(x) = cx+1-c^2$