A little thought

For this we need to know two things:

‣ Lambert series

‣ Binets formula

Using binets formula we know $F_n=\frac{\phi^n-(1-\phi)^n}{\sqrt{5}}$

$S= \sum_{n=1}^{\infty} \frac{\varphi(n)(1-\phi)^{n}}{F_{n}}= \sqrt{5} \sum_{n=1}^{\infty} \frac{\varphi(n)(1-\phi)^{n}}{\phi^n-(1-\phi)^n}= \sqrt{5}\sum_{n=1}^{\infty} \frac{\varphi(n)\left(\frac{1-\phi}{\phi}\right)^{n}}{1-\left(\frac{1-\phi}{\phi}\right)^{n}}$

Now we know that the lambert series for $φ(x)$ where $|q|<1$ is $\sum_{n=1}^{\infty}\frac{φ(n)q^n}{1-q^n}=\frac{q}{(1-q)^2}$

Note $|\frac{1-\phi}{\phi}|<1$

$\therefore S=\sqrt{5}\frac{\frac{1-\phi}{\phi}}{\left(1-\frac{1-\phi}{\phi}\right)^2}=\sqrt{5}\frac{\phi(1-\phi)}{4\phi^2-4\phi+1}=-\frac{\sqrt{5}}{5}$

We can then evaluate the answer to be $\left(-\frac{\sqrt{5}}{5}\right)^{-2}=\boxed{5}$