How to solve a functional equation involving log?

Question

It's given that $$f(xy)=\frac {f (x)}{y}+\frac {f (y)}{x}$$ Also $x,y>0$ and $f(x)$ is differentiable for $x>0$ such that $f(e)=\frac{1}{e}$. By the look of the functional equation I am sure it does involve log at some point . By common substitutions I have been able to deduce that f(1)=0 and f(1/e)=-e but I am not sure how to proceed . Any hint is appreciated.

Answer 1

Multiplying with $xy$ yields $$(xy)f(xy) = xf(x) + yf(y).$$ Define $g(x) := xf(x)$ to get $g(xy) = g(x) + g(y)$. Dou you know a/the function that satisfies this?

Answer 2

Let $g(x)=xf(x)$ so we have $$g(xy) = g(x)+g(y)$$

Let $h(x)=g(e^{x})$, then $$h(x+y) = g(e^{x+y})=g(e^x\cdot e^y)= g(e^x)+g(e^y)=h(x)+h(y)$$

So $h$ is Cauchy function and since it is differentiable it is linear, so $h(x)=ax$ for some real $a$.

Since $h(1)= g(e)= ef(e)= e{1\over e} = 1$ so $a=1$ and thus $$x = g(e^x)\implies g(x) = \log x$$