Solution to a puzzle by @Numerus_A

Assume $x \in [0,2a]$. Because $\f{a,b}{x} \geq 0$, we may rewrite

\begin{equation} \f{a,b}{\g{a,b}{x}} = \sqrt{\f{a,b}{\g{a,b}{x}}^2} = \sqrt{2}\sqrt{a^2+b^2+\g{a,b}{x}^2 - \sqrt{(a^2+b^2+\g{a,b}{x}^2)^2 - 4a^2\g{a,b}{x}^2}} \end{equation}

\begin{equation} \g{a,b}{x}^2 = \frac{{\left(4 \, a^{2} + 4 \, b^{2} - x^{2}\right)} x^{2}}{4 \, {\left(2 \, a + x\right)} {\left(2 \, a - x\right)}} \end{equation}

\begin{equation} a^2 + b^2 + \g{a,b}{x}^2 = \frac{16 \, a^{4} + 16 \, a^{2} b^{2} - x^{4}}{4 \, {\left(2 \, a + x\right)} {\left(2 \, a - x\right)}} \end{equation}

\begin{equation} \sqrt{\left(a^2 + b^2 + \g{a,b}{x}^2\right)^2 - 4a^2\g{a,b}{x}^2} = \frac{{16 \, a^{4} + 16 \, a^{2} b^{2} - 8 \, a^{2} x^{2} + x^{4}}}{4 \, {\left(2 \, a + x\right)} {\left(2 \, a - x\right)}} \end{equation}

Plugging in these values into the above expression for $\f{a,b}{\g{a,b}{x}}$ yields

\begin{equation} \f{a,b}{\g{a,b}{x}} = \sqrt{2}\sqrt{\frac{8a^2x^2-2x^4}{4 \, {\left(2 \, a + x\right)} {\left(2 \, a - x\right)}}} = x \end{equation}

We can use Binet-type expressions for Lucas and Fibonacci numbers:
\begin{align} 5F_{2n}^2 + 4 &= 5\left( \frac{\phi^{2n} - \phi^{-2n}}{\sqrt{5}} \right)^2 + 4\\ &= \phi^{4n} - 2 + \phi^{-4n} + 4 \\ &= \phi^{4n} + 2 + \phi^{-4n} \\ &= \left( \phi^{2n} + \phi^{-2n} \right)^2 \\ &= L_{2n}^2 \end{align}
Alternatively we can also compute in the $\mathbb{Q}$-algebra $\mathbb{Q}[\frac{1}{2},\frac{\sqrt{5}}{2}]$, as in a previous tweet by @Numerus_A :
\begin{align} 1 &= \phi^{-2n}\phi^{2n} \\ &= \frac{L_{2n} - F_{2n}\sqrt{5}}{2} \frac{L_{2n} + F_{2n}\sqrt{5}}{2} \\ &= \frac{L_{2n}^2 - 5F_{2n}^2}{4} \end{align}