Functional equation, find particular value given $f(ab)=bf(a)+af(b)$

Let $f(x)$ be a function such that $f(ab) = bf(a) + af(b)$ for all nonzero real numbers. Given that $f(4) = 3$, which of the following is a possible value of $f(2018)$?

(A) $0\quad$ (B) $\dfrac34\quad$ (C) $\dfrac43\quad$ (D) $1512\quad$ (E) $2688\quad$

By substitution, it's easy to find that $f(1) = 0$ and $f(2) = \dfrac34$. But how can I get to $f(2018)$?