Functional Equation

Plug $y=1$ : $f(f(x)^2) = x^3f(x)$ ...... $(1)$

$\Rightarrow \frac{x^2}{f(f(x)^2)} = \frac{1}{xf(x)} = k$ (say)

Then, $f(x) = \frac{1}{kx}$

Put this in $(1)$ . We have, $f(f(x)^2) = kx^2$ and, $x^3f(x) = \frac{x^2}{k}$

Equating them, we get: $k^2 = 1 \Rightarrow k = 1,-1$

So, $f(x) = \frac{1}{x}, \frac{-1}{x}$ . But, $f: \mathbb {Q}^+ \rightarrow \mathbb {Q}^+$ , which is not satisfied by $f(x) = \frac{-1}{x}$ . Thus, $f(x) = \frac{1}{x}$ . Put this in the parent equation to check whether it satisfies. And it does. Then, $f(\frac{1}{2016}) = 2016$

Putting $y=1$ , then $f(f(x)^2) = x^3f(x)$ . Let the degree of $f(x)$ be $n$ . The the degree of the LHS is $2n^2$ and that of the RHS is $n+3$ . Equating the degrees on both sides, we have:

$\begin{aligned} 2n^2 & = n+3 \\ 2n^2 - n -3 & = 0 \\ (2n-3)(n+1) & = 0 \end{aligned}$

This means that $n = \frac 32$ or $n=-1$ . Since $f: \mathbb {Q^+ \to Q^+}$ , $\implies n = -1$ . $\implies f(x) = \frac 1x$ . Since $f: \mathbb {Q^+ \to Q^+}$ , $f(x) \ne = \frac 1x$ . Therefore, $f\left(\frac 1{2016}\right) = \boxed{2016}$ .