Functional Equation!

Substitute $x \rightarrow n+1, y \rightarrow n$ to get $\dfrac{f(n+1)^2}{f(n)}+f(n+1)=f(n+2)$ $\dfrac{f(n+1)}{f(n)}+1=\dfrac{f(n+2)}{f(n+1)}$ Define another sequence $g(n)=\dfrac{f(n)}{f(n-1)},g(1)=1$ $g(n+2)=g(n+1)+1,g(1)=1$ $\Rightarrow g(n)=n$ $f(n)=nf(n-1)$ $\Rightarrow f(n)=n!$ Hence $\mbox{Index of 3}=672+224+74+24+8+2=\boxed{1004}$

Suppose $x > 0$ and let $y = x-1$ to get $\frac{f(x)^2}{f(x-1)} + f(x) = f(x+1)$ . I claim that $f(x) = x!$ . To prove this, we use strong induction on $x$ . The base case is trivial, not assume the claim is true for $x = k$ . We have $f(k+1) = \frac{f(k)^2}{f(k-1)} + f(k) = k \cdot k! + k! = (k+1)!$ and we are done. So, it suffices to find the exponent of $3$ in the prime factorization of $2016!$ , which is just $\left \lfloor \frac{2016}{3} \right \rfloor + \left \lfloor \frac{2016}{3^2} \right \rfloor + \left \lfloor \frac{2016}{3^3} \right \rfloor + \left \lfloor \frac{2016}{3^4} \right \rfloor + \left \lfloor \frac{2016}{3^5} \right \rfloor + \left \lfloor \frac{2016}{3^6} \right \rfloor = \boxed{1004}$