The summation $\sum_{k=1}^{\infty} \frac{\phi^{-2k}}{k^2}=\text{Li}_2(1/\phi^2)=\frac{\pi^2}{15}-\ln^2 \phi$

Question

In a recent post

Computing $\int_0^{1/2}\frac{\sinh^{-1}(u)}{u} \,du=\frac{\pi^2}{20}$, $\zeta(2)=\frac53 \sum_{n=0}^\infty \frac{(-1)^n\binom{2n}{n}}{2^{4n}(2n+1)^2}$

I evaluated the required Integral using a very interesting result that $$\sum_{k=1}^{\infty} \frac{\phi^{-2k}}{k^2}=\text{Li}_2(1/\phi^2)=\frac{\pi^2}{15}-\ln^2 \phi,~~ \phi=\frac{1+\sqrt{5}}{2}\qquad(*)$$ See https://en.wikipedia.org/wiki/Polylogarithm#Relationship_to_other_functions

The question is: How to prove (*)?