The normalized Fermat curve is $X^n+Y^n=1$. We have of course infinite possibilities for parametrisation, but the periodicity is a special characteristic here.
E.g. $\cos^2x+\sin^2x=1$ has periodicity, but not $(\frac{1-x^2}{1+x^2})^2+(\frac{2x}{1+x^2})^2=1$.
Here I am interested in periodic functions for $n:=3$. Be $x\in \mathbb{C}$.
Be $f^\prime(x)=-\frac{g^2}{f}(x)$ (note: pole $x_0$ for $f(x_0)=0$) , $g^\prime(x)=f(x)$, $f(0)=1$ and $g(0)=0$.
=> $f^3(x)+g^3(x)=1$
Be $\eta$ and $\mu$ such that $f(\eta)=0,\, g(\eta)=1$ and $f(\mu)=g(\mu)=\frac{1}{\sqrt[3]{2}}$.
$X^{-1}$ here is used as inverse function for $X$ within a valid value range. After a few transformations one get e.g.
$(\frac{g}{f})^{-1}(x)=\int\limits_0^x \frac{dt}{1+t^3}$,
$g^{-1}(x)=\int\limits_0^x \frac{dt}{\sqrt[3]{1+t^3}}$,
$f^{-1}(x)=\int\limits_x^1 \frac{tdt}{\sqrt[3]{1+t^3}^2}$,
$\mu=(\frac{g}{f})^{-1}(1)=\frac{\pi}{3\sqrt{3}}+\frac{1}{3}\ln 2$,
$\eta=\frac{\Gamma(\frac{1}{3})\Gamma(\frac{2}{3})}{\Gamma(1)}=\frac{2\pi}{3\sqrt{3}}=2\mu-\frac{1}{3}\ln 4$.
Be $\lambda:=e^{i\frac{2\pi}{3}}$.
Note: With the Taylor expansion around 0 for $f(x)$ and $g(x)$ one get $f(\lambda x)=f(x)$ and $g(\lambda x)=\lambda g(x)$.
The addition theorems for $g$ and $f$.
With $(\frac{g}{f})^{-1}(x)=\frac{1}{3}(\ln(1+x)+\lambda^2\ln(1+\lambda x)+\lambda\ln(1+\lambda^2 x))$ follows $(\frac{g}{f})^{-1}(x)+(\frac{g}{f})^{-1}(y)+\frac{1}{3}\ln(1-\frac{3xy}{(1+x)(1+y)})=(\frac{g}{f})^{-1}(\frac{x+y-xy}{1-xy})$
and with $x\to\frac{g}{f}(x)$, $y\to\frac{g}{f}(y)$ and $f^3(z)+g^3(z)=1$ one get
$f(x+y+\frac{1}{3}\ln(1-\frac{3g(x)g(y)}{(f(x)+g(x))(f(y)+g(y))}))=$ $\frac{f(x)f(y)-g(x)g(y)}{\sqrt[3]{(f(x)f(y)-g(x)g(y))^3+(g(x)f(y)+f(x)g(y)-g(x)g(y))^3}}$ and
$g(x+y+\frac{1}{3}\ln(1-\frac{3g(x)g(y)}{(f(x)+g(x))(f(y)+g(y))}))=$ $\frac{g(x)f(y)+f(x)g(y)-g(x)g(y)}{\sqrt[3]{(f(x)f(y)-g(x)g(y))^3+(g(x)f(y)+f(x)g(y)-g(x)g(y))^3}}$
Be $\phi:=\eta+\mu-\frac{1}{3}\ln(-2)=\frac{\pi}{\sqrt{3}}-\frac{\ln(-1)}{3}$.
With $x:=\eta$ and $y:=\mu$ one get $g(\phi)=0$ and $f(\phi)=1$.
General: $g(a)=0$ => $g(x+a)=g(x)$ , here: $\Re(a)\ge 0$.
$x:=\lambda a$ and $y:=\lambda^2 a$: $0=g(0)=g(-a)$ , one get $g(\pm a)=0$
QUESTION: How can we interpret $\ln(-1)$ here ?
Does it mean that $g(x+k\cdot\frac{\pi}{\sqrt{3}}+l\cdot\frac{i\pi}{3})=g(x)$ ? $k,l\in\mathbb{Z}$, maybe only $k\in\mathbb{N_0}$
Does $l$ depend on $k$ or are they independent of each other ?
A supplementary question: Has $\frac{g(\phi x)}{\phi x}$ anything to do with $\prod\limits_{k=1}^\infty (1-(\frac{x}{k})^3)$ ?
EDIT:
Because of the comments below, the double periodicity here has the form
$g(x+k(\frac{\pi}{\sqrt{3}}+i(2l+1)\frac{\pi}{3}))=g(x)$
with $k,l\in\mathbb{Z}$.
