Function with derivative-like property: $f(ab) = af(b) + bf(a)$ Let $f$ be a function $(0;+\infty)\to\mathbb{R}$ with following property:
$f(ab) = af(b) + bf(a)$.
What can $f$ be?
It can be seen that functions $\delta_p(x) = px\space ln(x)$ work.
Now, let $f_0$ be a solution and $f_0(x_0)=y_0$, $x_0\neq1$. Because $\delta$ is continuous by $p$, there exists such $p_0$ that $\delta_{p_0}(x_0)=y_0$.
It can be proven that
$\forall r\in\mathbb{Q}\space f_0(x_0^r)=\delta_{p_0}(x_0^r)$.  
In other words, any solution must be a union of several $\delta_{p_S}$ functions restricted to subsets $Q_S\subset\mathbb{R}^+$, which are equivalence classes of relation $a\sim b$ if $a = b^q, q\in\mathbb{Q}$.
The question is, can a solution include two or more different $\delta_p$, i.e. can it be non-continuous, i.e. can it be not of form $\delta_p$?
 A: Substitute $f(x)=xg(\ln x)$ (so $g(y)=e^{-y}f(e^y)$). Then $g\colon \mathbb R\to\mathbb R$ satisfies the Cauchy's equation $g(a+b)=g(a)+g(b)$, whose solutions are described in a known way (and may well be discontinous).
A: As $\pi$r8 pointed out in the comments, this functional equation is really close to Cauchy's. 
If $f$ is a solution and that you denote $g:x\rightarrow f(e^x)e^{-x}$, then for all $x,y$ $g(x+y) = e^{-x+y} (e^y f(e^x) + e^x f(e^y))$, $\textit{i.e.}$ $g(x+y) = g(x) + g(y)$, which is Cauchy's functional equation.
Thus, if you suppose for instance that $f$ is continuous at one point at least, you can prove that $g(x) = a x$ for some $a$, and thus $f = \delta_a$ is one of the solutions you identified.
But $f$ can indeed be non-continuous, because you can find $g$ non-continuous satisfying $g(x+y) = g(x) + g(y)$.
Classically, in $\mathbb{R}$ considered as a vector space over $\mathbb{Q}$, there exists a subspace $B$ s.t. $\mathbb{R} = \mathbb{Q} \oplus B$. We can consider $g$ the projection along $B$ onto $\mathbb{Q}$.
As $g(\mathbb{R}) = \mathbb{Q}$, you know that $g$ is nowhere continuous. And we define, for $x>0$, $f(x) = xg(\ln (x))$. Then for $x,y>0,\ f(xy) = xy\ g(\ln(x)+\ln(y))$ so $$f(xy) = yx\ g(\ln(x)) + xy\ g(\ln(y)) = yf(x)+xf(y)$$
But $f$ is not one of the $\delta_a$ (f is not continuous, not bounded, ...)
A: First, verify that $f(0) = f(1) = 0$
For the functions that have up to third derivative.
$f(xy) = xf(y)+yf(x)$, differentiate by $x$ two times and by $y$ one time:
$$
yf'(xy)=f(y)+yf'(x) $$$$ y^2f''(xy)=yf''(x) $$$$ yf''(xy)=f''(x)
$$
$$
f''(xy)+xyf'''(xy)=0
$$
Now let $xy=z, f''(z) = \psi(z)$ and we gets differential equation on $\psi$: $\psi(z)+z\psi'(z)=0$ so that $\psi(z) = \frac{C1}{z}$
Then $f'(z)=C1\ln(z)+C2$ and $f(z) = C3 + C2z + C1(z\ln(z) - 1)$
Considering initial condition: $C3-C1=0, C2+C3-C1=0$ so $C2=0$ your initial form is the only correct.
