What's the value of this Viète-style product involving the golden ratio?

What you're basically looking for is a function $f(x)$ such that $f(2x)=f^2(x)-1$ and $f(0)=\phi$, from there: \begin{align} 2f'(2x)&=2f(x)f'(x)\\\\ \frac{f'(2x)}{f'(x)}&=f(x)\\\\ \frac{f'(x)}{f'(x/2)}&=f(x/2)\\\\ \frac{f'(x)}{f'(x/2^n)}&=\prod_{k=1}^n f(x/2^k) \end{align} and, given a value $x_0$ such that $f(x_0)=1$, \begin{align} \Phi&=\prod_{k=1}^{\infty} \frac{f(x_0/2^k)}{\phi}\\\\ &=\lim_{n\rightarrow\infty}\phi^{-n} \prod_{k=1}^n f(x_0/2^k)\\\\ &=\lim_{n\rightarrow\infty}\phi^{-n} \frac{f'(x_0)}{f'(x_0/2^n)}\\\\ &=\lim_{h\rightarrow0}h^\alpha \frac{f'(x_0)}{f'(hx_0)} \end{align}

where $\alpha=\frac{\ln(\phi)}{\ln(2)}$. Unfortunately, I have no idea how to get $f(x)$, and the fact that $f(x)=1+O(x^{1+\alpha})$ does not make finding this function look easy.