<s>[INST] 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.
 [/INST] 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
$$f(2015) = \frac{2015}{5}\cdot 3 = 1209$$
</s>