How to show that? $$S=\sum_{n=0}^{\infty}(-1)^{n+1}\left(\frac{F_n}{F_{n+1}F_{n+2}}\right)^2=\frac{1}{\phi^3}$$
Where $F_n$ Fibonacci number
$F_n=\frac{\phi^n-(-\phi)^{-n}}{\sqrt{5}}$
$$F_n^2=\frac{\phi^{2n}-2\phi^n(-\phi)^{-n}+(-\phi)^{-2n}}{5}$$
$$S=\sum_{n=1}^{\infty}(-1)^{n+1}2\left(\frac{\phi^n}{F_{n+1}F_{n+2}}\right)^2+\sum_{n=1}^{\infty}\left(\frac{1}{F_{n+1}F_{n+2}}\right)^2$$
Lemma: Let $F_n$ be Fibonacci Numbers and $L_n$ be Lucas Numbers, then $$ F_nF_{n+1}-F_{n-k}F_{n+k+1}=\frac{(-1)^{n+1}+(-1)^{n-k}L_{2k+1}}5\tag1 $$ Proof: Using $F_n=\frac{\phi^n-(-1/\phi^n)}{\sqrt5}$ and $L_n=\phi^n+(-1/\phi)^n$, we get $$ \begin{align} &5(F_nF_{n+1}-F_{n-k}F_{n+k+1})\\ &=\left(\phi^n-(-1/\phi)^n\right)\left(\phi^{n+1}-(-1/\phi)^{n+1}\right) -\left(\phi^{n-k}-(-1/\phi)^{n-k}\right)\left(\phi^{n+k+1}-(-1/\phi)^{n+k+1}\right)\\ &=(-1)^{n+1}(\phi-1/\phi)+(-1)^{n-k}\left(\phi^{2k+1}-1/\phi^{2k+1}\right)\\ &=(-1)^{n+1}+(-1)^{n-k}L_{2k+1} \end{align} $$ $\large\square$
Therefore, $$ \begin{align} \sum_{k=0}^\infty\frac1{F_{2k+1}F_{2k+3}} &=\sum_{k=0}^\infty\frac{F_{2k+1}F_{2k+2}-F_{2k}F_{2k+3}}{F_{2k+1}F_{2k+3}}\tag{2a}\\ &=\sum_{k=0}^\infty\left(\frac{F_{2k+2}}{F_{2k+3}}-\frac{F_{2k}}{F_{2k+1}}\right)\tag{2b}\\ &=\frac1\phi\tag{2c} \end{align} $$ Explanation:
$\text{(2a)}$: apply the Lemma substituting $(n,k)\mapsto(2k+1,1)$
$\text{(2b)}$: algebra
$\text{(2c)}$: the partial sum to $n$ telescopes to $\frac{F_{2n+2}}{F_{2n+3}}$, which limits to $\frac1\phi$
$$ \begin{align} \sum_{n=0}^\infty(-1)^{n+1}\left(\frac{F_n}{F_{n+1}F_{n+2}}\right)^2 &=\sum_{n=0}^\infty(-1)^{n+1}\left(\frac{F_{n+2}-F_{n+1}}{F_{n+1}F_{n+2}}\right)^2\tag{3a}\\ &=\sum_{n=0}^\infty(-1)^{n+1}\left(\frac1{F_{n+1}^2}+\frac1{F_{n+2}^2}-\frac2{F_{n+1}F_{n+2}}\right)\tag{3b}\\ &=\sum_{n=0}^\infty(-1)^{n+1}\frac1{F_{n+1}^2}-\sum_{n=1}^\infty(-1)^{n+1}\frac1{F_{n+1}^2}\\ &+2\sum_{n=0}^\infty\left(\frac1{F_{2n+1}F_{2n+2}}-\frac1{F_{2n+2}F_{2n+3}}\right)\tag{3c}\\ &=-1+2\sum_{n=0}^\infty\frac1{F_{2n+1}F_{2n+3}}\tag{3d}\\[3pt] &=-1+\frac2\phi\tag{3e}\\[6pt] &=\frac1{\phi^3}\tag{3f} \end{align} $$ Explanation:
$\text{(3a)}$: $F_n=F_{n+2}-F_{n+1}$
$\text{(3b)}$: algebra
$\text{(3c)}$: break the sum into three pieces
$\phantom{\text{(3c):}}$ substitute $n\mapsto n-1$ in the second piece
$\phantom{\text{(3c):}}$ break the third piece into two to remove $(-1)^{n+1}$
$\text{(3d)}$: cancel identical terms in the first two sums
$\phantom{\text{(3d):}}$ combine terms in the third sum using $F_{2k+3}-F_{2k+1}=F_{2k+2}$
$\phantom{\text{(3d):}}$ and cancel $F_{2k+2}$ in the numerator and denominator
$\text{(3e)}$: apply $(2)$
$\text{(3f)}$: $2-\phi=\frac1{\phi^2}$