If $f(4xy)=2y[f(x+y)+f(x-y)]$ and $f(5)=3$, find $f(2015)$ 
Suppose the function $f:\Bbb R\to\Bbb R$ satisfies the following conditions:
$$\begin{align}
f(4xy)&=2y[f(x+y)+f(x-y)] \\[4pt]
f(5)&=3
\end{align}$$
Find the value of $f(2015)$.

I have tried to find some other hiding condition, like $f(0)=0,$ but which is useless.
 A: With the provided function,
$$f\left(4xy\right) = 2y\left[f\left(x + y\right) + f\left(x - y\right)\right] \tag{1}\label{eq1}$$
First, substitute $y = 0$ to get
$$f\left(0\right) = 0 \tag{2}\label{eq2}$$
Next, substitute $x = 0$ and use \eqref{eq2} to get
$$0 = 2y\left[f\left(y\right) + f\left(-y\right)\right] \tag{3}\label{eq3}$$
Thus, for all values of $y$ other than $0$, dividing both sides by $2y$ gives
$$f\left(y\right) = -f\left(-y\right) \tag{4}\label{eq4}$$
In other words, $f$ is an odd function.  Next, substituting $y = 1$ into \eqref{eq1} gives
$$f\left(4x\right) = 2\left[f\left(x + 1\right) + f\left(x - 1\right)\right] \tag{5}\label{eq5}$$
Now, using $x = 1$ in \eqref{eq5} gives
$$f\left(4\right) = 2\left[f\left(2\right) + f\left(0\right)\right] \tag{6}\label{eq6}$$
Thus, using \eqref{eq2} gives
$$f\left(4\right) = 2f\left(2\right) \tag{7}\label{eq7}$$
Similarly, using $x = 3$ in \eqref{eq5} gives
$$f\left(12\right) = 2\left[f\left(4\right) + f\left(2\right)\right] \tag{8}\label{eq8}$$
Thus, using \eqref{eq7} gives
$$f\left(12\right) = 6f\left(2\right) \tag{9}\label{eq9}$$
Note that \eqref{eq2}, \eqref{eq4}, \eqref{eq7} and \eqref{eq9} all satisfy
$$f\left(nx\right) = nf\left(x\right) \forall \; n \in N \tag{10}\label{eq10}$$
In particular, for \eqref{eq2}, $n = 0 \; \forall \; x \in R$; for \eqref{eq4}, $n = -1 \; \forall \; x \in R$; and for \eqref{eq7} & \eqref{eq9}, $x = 2$ only, with $n = 2$ & $n = 6$, respectively. This doesn't prove \eqref{eq10} always works, but it suggests that $f$ is a linear multiple of $x$, i.e, $f\left(x\right) = kx \text{ for a non-zero constant } k \in R$, which from the given condition of
$$f\left(5\right) = 3 \tag{11}\label{eq11}$$
gives a value of $k = \frac{3}{5}$ so the function would be
$$f\left(x\right) = \cfrac{3x}{5} \tag{12}\label{eq12}$$
To confirm this, substitute \eqref{eq12} into \eqref{eq5} to give on the left side
$$\cfrac{12x}{5} \tag{13}\label{eq13}$$
with the right side becoming
$$2\left[\cfrac{3\left(x + 1\right)}{5} + \cfrac{3\left(x - 1\right)}{5} \right] = 2\left[\cfrac{\left(3x + 3 + 3x - 3\right)}{5} \right] = \cfrac{12x}{5} \tag{14}\label{eq14}$$
Similarly, using \eqref{eq12} in \eqref{eq1} gives a left & right hand side of $\frac{12xy}{5}$, confirming it's also a solution of the original equation.  I'm not quite sure offhand how to prove it's unique, but I assume it is.  We thus get a final answer of
$$f\left(2015\right) = \cfrac{3 \times 2015}{5} = 1209 \tag{15}\label{eq15}$$
Note: The solution could involve just "guessing" the function being a constant multiple of the argument, and then showing this works to determine the final answer, but I thought it might be useful for people to see how one can approach solving this problem somewhat systematically as I did. In particular, I tried using small constant values for $x$ and $y$, checking what information this offered, and then seeing if I could leverage this when using other values. For example, I started with $x = 0$ and $y = 0$, then used $y = 1$, $x = 1$ and $x = 3$. Each time, I used previous details to help simplify and learn more about the new values, until I saw the pattern which I then confirmed worked.
A: Firstly, it’s trivial that
$$f(0) = f(4\cdot 0\cdot 0) = 2\cdot 0 \cdot [f(0) + f(0)] = 0$$
Next, we see that
$$0 = f(0) = f(4\cdot 0 \cdot y) = 2y\cdot [f(y) + f(-y)]\text{,}$$
which implies that $f(-y) = -f(y)$ for all $y\neq 0$.
Afterwards, one notices that
$$f(4xy) = f(4yx)$$
Hence,
$$2y\cdot [f(x+y) + f(x-y)] = 2x\cdot [f(x+y) + f(y-x)]$$
$$\Leftrightarrow (x-y)\cdot f(x+y) = (x+y)\cdot f(x-y)$$
$$\Leftrightarrow f(x+y) = \frac{x+y}{x-y}\cdot f(x-y)$$
If we substitute now $x=1010$ and $y=1005$, we get that
$$f(2015) = \frac{2015}{5}\cdot 3 = 1209$$
