Function with derivative-like property: $f(ab) = af(b) + bf(a)$

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).