Let $\psi$ be an elliptic function with periodic lattice $\mathbb{Z}[\omega]= \mathbb{Z}\omega \oplus \mathbb{Z}$, a pole of order $2$ at $0$, and simple zeros at $\pm\dfrac{\omega-1}{3}$. Here $\omega= \exp(2 \pi i/3)$.
Assume that $\psi(1/3)=1$.
Why is it true that $\psi\left( \frac{\omega}{3} \right) \psi\left( \frac{2-\omega}{3}\right)= 1? $
You can check that $\psi(\omega z)/\psi(z)$ is $\Bbb Z[\omega]$-periodic and has no zero nor pole, so it is a constant.
Near $z = 0$, $\psi(z) \sim a/z^2$ for some nonzero complex number $a$, and so $\psi(\omega z) \sim a/\omega^2 z^2 = \omega a/z^2$, hence $\psi(\omega z)= \omega \psi(z)$, forall $z \in \Bbb C$.
Applying this repeatedly to $z=1/3$ gives you $\psi(\omega/3) = \omega$ and $\psi((2-\omega)/3) = \omega^2$.