I try to solve for $f:(0,\infty)\rightarrow (0,\infty)$, s.t. for all $x,y>0$:
$$f(xf(y))=f(xy)+x.$$
By now, I tried a lot of substitutions like $x=y=1, x=1/y$ etc. But until now I could not find the solution(s). Does anyone has a hint for me how to attack this problem?
Replacing $x$ with $f(x)$ in the functional equation, we find that $$ f(f(x) f(y)) = f(f(x)y) + f(x) = f(xy) + y + f(x). $$
Swapping $x$ and $y$ in this relation also gives us that $$ f(f(x) f(y)) = f(xy) + x + f(y). $$
Thus we have that $$ f(xy) + y + f(x) = f(xy) + x + f(y) $$ for all $x$ and $y$, and so we have that $$ x + f(y) = y + f(x) $$ for all $x$ and $y$. Now let $y = 1$. Then we get that $$ f(x) = x + f(1) - 1 $$ for all $x > 0$. In other words, there is some constant $c$ such that $f(x) = x + c$ for all $x > 0$. Substituting this into the original equation shows us that we must have that $c = 1$.