Given the Wallis's product of $\pi$,
$${\pi\over 2}=\prod_{n=1}^{\infty}{4n^2\over 4n^2-1}=\prod_{n=1}^{\infty}\left(1+{1\over 4n^2-1}\right)\tag1$$
$${1\over 2}=\sum_{n=1}^{\infty}{1\over 4n^2-1}\tag2$$
$(2)$ is a telescope sum.
We can rewrite $(1)$ in term of $\pi$
$$\pi={\prod_{n=1}^{\infty}\left(1+{1\over 4n^2-1}\right)\over \sum_{n=1}^{\infty}{1\over 4n^2-1}}\tag3$$
Sondow called $(3)$ as symmetric formula for $\pi$.
We observed $(3)$ and was able to generalised to:
$$\pi^{F_k}={\prod_{n=1}^{\infty}\left(1+{1\over 4n^2-1}\right)^{(-1)^{n-1}F_{k+1}+F_{k+2}}\over \left(\sum_{n=1}^{\infty}{1\over 4n^2-1}\right)^{-F_{k-1}}}\tag4$$
Let $F_0=0$ and $F_1=1$
Where $F_{k+1}=F_{k-1}+F_k$ and negative Fibonacci numbers are:
$F_{-1}=1, F_{-2}=-1, F_{-3}=2, F_{-4}=3, F_{-5}=-5$ and so on ...
Setting $k=-1$ into $(4)$ we arrived at $(3)$
Note: We arrived at $(4)$ via a mean of trial and error (experimental-mathematics) approach but we don't know how to prove it mathematically.
How can we prove or disprove $(4)$?
The result you mention is just plain algebraic manipulation and equations $(1),(2)$. I will mention only an outline of the manipulation needed. When we raise the equation $(3)$ to power $F_{k} $ then both numerator and denominator get raised to power $F_{k} $. For denominator express the power as $F_{k+1}-F_{k-1}$ and thus using $(2)$ you get the denominator as $2^{-F_{k+1}}$ times denominator of $(4)$. The power in numerator is expressed as $F_{k+2}-F_{k+1} $. Comparing with equation $(4)$ we can see that we only need to establish the equation $$2=\prod\left(1+\frac{1}{4n^{2}-1}\right) \prod\left(1+\frac{1}{4n^{2}-1}\right)^{(-1)^{n-1}}$$ ie $$2=\prod\frac{4n^{2}}{4n^{2}-1}\cdot\frac{4(2n-1)^{2}}{4(2n-1)^{2}-1}\cdot\frac{4(2n)^{2}-1}{4(2n)^{2}}$$ which is just plain algebraic manipulation (if you evaluate the product for $n=1$ to $N$ you will get $(4N+1)/(2N+1)$).
Note that actual values of Fibonacci numbers do not matter only the equation $F_{k+1}=F_{k} +F_{k-1}$ is important.