If Only It Was (n-2)(n+2)

Note that $F_{n-2}+F_{n+2}=3F_n$ $\Rightarrow \sum_{n=3}^\infty \dfrac{1}{F_{n-2}F_{n+2}}$ $=\sum_{n=3}^\infty \left(\dfrac{1}{F_{n-2}F_{n+2}} \times \dfrac{F_{n-2}+F_{n+2}}{3F_n}\right)$ $=\dfrac13 \sum_{n=3}^\infty \left(\dfrac{1}{F_{n-2}F_n}+\dfrac{1}{F_n F_{n+2}}\right)$ $=\dfrac13 \left[\dfrac{1}{F_1 F_3}+\dfrac{1}{F_2 F_4}+2\sum_{n=3}^\infty \dfrac{1}{F_n F_{n+2}}\right]$ $=\dfrac{5}{18}+\dfrac23 \left [\sum_{n=3}^\infty \dfrac{1}{F_n F_{n+2}} \times \dfrac{F_{n+2}-F_n}{F_{n+1}}\right]$ $=\dfrac{5}{18}+\dfrac23 \sum_{n=3}^\infty \left(\dfrac{1}{F_n F_{n+1}}-\dfrac{1}{F_{n+1} F_{n+2}}\right)$ $\dfrac{5}{18}+\dfrac23 \times \dfrac{1}{F_3 F_4}$ $=\boxed{\dfrac{7}{18}}$

We can proceed by dividing and multiplying by Fn and then opening it till we get terms of Fn ,Fn+2,Fn-2 . Then we see that a term same as what we want is generated, so taking it on other side and adding makes it 3S. Now simple telescoping technique yields our answer.