Nice!! Observe that $ f(n+1)f(n-1)-f^2(n)=-(xy)^{n-1}$ and so it follows that $ (xy)^n$ and $ (xy)^{n+1}$ are integers. Thus xy is rational and algebraic integer and so it is integer. Next, observe that you can write $f(n)$ as a polynomial with integer coefficients in xy and x+y (or, at least, it seems it is true by cheking small cases, so it must be true ) such that $ (x+y)$ appears with maximal power alone (not multiplied with xy) and with coefficient 1. It thus follows that x+y is algebraic integer and rational (since f(n), f(n+1), f(n+2) and xy are integers) and thus it is integer. The conclusion follows. This, if the claim with the polynomial is true.

Sorry, I was idiot, the assertion with the polynomial comes by induction from the fact that $ f(n+1)=(x+y)f(n)-xyf(n-1)$ and xy integer. The assertion is: for all n, f(n) has the form $ f(n)=(x+y)^{n-1}+P(x+y,xy)$ where P is a polynomial with integer coefficients and such that all powers of x+y has exponent at most n-2.

This was also used in the 1995 Bulgarian olympiad.