How to prove: $f(x)$ is differentiable on $(0,+\infty)$ The function $f(x)$ is defined on    $(0,+\infty)$. We know $f'(1)$ exists and we have that $$\forall x,y \in(0,+\infty), \quad  f(xy)=yf(x)+xf(y)$$ How to prove:$f(x)$ is differentiable on $(0,+\infty)$ and$$f'(x)=\frac{f(x)}{x}+f'(1)?$$
I've got $f(1)=0$, but have no idea to prove other things.
 A: Write $f(x):= x\>g(x)$ with a new unknown function $g$, defined on ${\mathbb R}_{>0}$. The function $g$ then satisfies the functional equation of the logarithm:
$$g(x\,y)=g(x)+g(y)\ ;\tag{1}$$
furthermore one has $g(1)=0$ and $g'(1)=1$. 
Using induction one deduces from $(1)$ that $g(x^n)=n\> g(x)$ for all $x>0$ and all $n\in{\mathbb N}_{\geq1}$. For any given $y>0$ we therefore have
$$g(y)=n\> g(y^{1/n})={y^{1/n}-1 \over{1/n}}\>{g(y^{1/n})-1 \over y^{1/n}-1}\qquad (n\geq 1)\ .\tag{2}$$
As
$$\lim_{t\to 0}{y^t-1\over t}=\log y\>, \qquad \lim_{x\to 1}{g(x)-g(1)\over x-1}=g'(1)=1$$
we conclude from $(2)$ that
$$g(y)=\log y\qquad(y>0)\ .\tag{3}$$
It follows that our problem has at most one solution, given by $(3)$. But it is easy to verify that the function $\log$ indeed fulfills all the requirements.  As a result, the function $f(x):=x\> g(x)$ solves the original problem, and is differentiable on all of ${\mathbb R}_{>0}$.
A: Using Dr. Blatter's hint, define $g:(0,\infty)\to\mathbb{R}$ by $g:x\mapsto\frac{f(x)}{x}$. So we have for every $x,y\in(0,\infty)$:
$$
g(xy)=\frac{f(xy)}{xy}=\frac{yf(x)+xf(y)}{xy}=\frac{yf(x)}{xy}+\frac{xf(y)}{xy}=\frac{f(x)}{x}+\frac{f(y)}{y}=g(x)+g(y).
$$
So $g(xy)=g(x)+g(y)$. Of course, we can immediately recognize one such function that will satisfy this relation, $\ln{x}$. Thus, one such $f$ that satisfies our hypotheses is $x\ln{x}$. In particular,
$$f'(x)=(x\ln{x})'=\ln{x}+\frac{x}{x}=\frac{x\ln{x}}{x}+\frac{1}{1}=\frac{f(x)}{x}+f'(1).$$
Indeed, any scalar multiple of $x\ln{x}$ will satisfy this relation.
