Question:
Let $f : \mathbb{R} \to \mathbb{R}$ be a function such that $f(5) = 3$ and
\[f(4xy) = 2y[f(x + y) + f(x - y)]\]for all real numbers $x$ and $y.$  Find $f(2015).$

Answer:
Setting $y = 0,$ we get $f(0) = 0.$

Then setting $x = 0,$ we get
\[f(0) = 2y[f(y) + f(-y)].\]Assuming $y \neq 0,$ we get $f(-y) + f(y) = 0.$  Hence, $f(-y) = -f(y)$ for all $y.$

We can reverse the roles of $x$ and $y$ to get
\[f(4xy) = 2x[f(x + y) + f(y - x)],\]so
\[2y[f(x + y) + f(x - y)] = 2x[f(x + y) + f(y - x)].\]Hence,
\[y f(x - y) - x f(y - x) = (x - y) f(x + y).\]Since $f(y - x) = -f(x - y),$
\[(x + y) f(x - y) = (x - y) f(x + y).\]We want to take $x$ and $y$ so that $x + y = 5$ and $x - y = 2015.$  Solving, we find $x = 1010$ and $y = -1005.$  Then
\[5 f(2015) = 2015 f(5),\]so $f(2015) = \frac{2015 f(5)}{5} = \boxed{1209}.$