Note that $f(k^2)=f(k) \cdot (k^2-k+1)$ so the inequality reduces to $2015 \ge \displaystyle\prod_{k=1}^{n} \frac{k^2+k+1}{k^2-k+1}$ Now note that since $(k+1)^2-(k+1)+1=k^2+k+1$ , the RHS telescopes and we get $2015 \ge n^2+n+1 \implies n \le \boxed{44}$

f(n^2)=n^4+n^2+1=(n^2+n+1)(n^2-n+1)

That is 2015 1^2 3^2 7^2….((n^2-n+1)^2 (n^2+n+1)≥1^2 3^2 7^2 …((n^2-n+1)^2 (n^2+n+1)^2

therefore 2015≥(n^2+n+1)^2

therefore n belongs to (-45.38 , 44.38)

since n is integer maximum value of it will be=44