Algebraic Functional Lemma

We are given: $\forall x,y\in Q^{+},f(f(x)^{2}y)=x^{3}f(xy).$ .(#)

Let us try two substitution:

(1) $x,y$ $\rightarrow$ $(x,\frac{1}{x})$ ,after which we get $f(\frac{f(x)^{2}}{x})$ = $x^{3}f(1)$ .①

(2) $x,y$ $\rightarrow$ $(x,\frac{1}{f(x)^{2}})$ ,after which we get $f(1)$ = $x^{3}f(\frac{x}{f(x)^{2}})$ .②

Now ①&② imply that $f(\frac{f(x)^{2}}{x})/f(\frac{x}{f(x)^{2}})$ = $x^{6}$ .③

Intending to use the symmetry,we take y such that:

$\frac{f(x)^{2}}{x}=\frac{y}{f(y)^{2}}$ .④

Substitute ④ into ③ we find: $x^{6}$ = $f(\frac{f(x)^{2}}{x})/f(\frac{x}{f(x)^{2}})$ = $f(\frac{y}{f(y)^{2}})/f(\frac{f(y)^{2}}{y})= \frac{1}{y^{6}}$

so $x^{6}y^{6}=1$ ,i.e. $xy=1$ .Take it into ④，we get $f(x)^{2}f(y)^{2}=xy=1$ ,and so $f(x)f(y)=1$ ,aha,see? We finally get $\large f(x)f(\frac{1}{x})=1,\forall x\in Q^{+}$

Since $f(x)$ cannot be constant 1 (easy to check in (#)) or x, it turns out that $f(x)=\frac{1}{x}$ and the answer is $\boxed{2018}$ .(i am sorry but i cannot state the last step clearly,and it's too late,i have to sleep...)

Guess $f(x)=1/x$

The left-hand side reduces to $\frac{1}{\frac{1}{x^2}y}$

The right-hand side reduces to $\frac{x^3}{\frac{1}{xy}}$

By inspection, our guess works. Since only one such function satisfies the given conditions, we have $f(x)=1/x \Rightarrow f(1/2018)=2018$