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A Question regarding the Weierstrass $\wp$-function of the hexagonal lattice

Playing with the Weierstrass $\wp$-function of the hexagonal (or triangular) lattice $\mathbb{T}$, $$ \wp'(z)^2 = 4 \wp(z)^3 - 1, $$ I noticed that the zeros of $\wp'(z) + \sqrt{3}$ are $$ \frac{\varpi}{3} + \omega, \quad e^{\pi i/3} \frac{2\varpi}{3} + \omega, \quad e^{2\pi i/3} \frac{\varpi}{3} + \omega, \qquad \omega \in \mathbb{T} $$ (clearly they are all simple), where $\varpi$ is the real period of $\wp$ given by $$ \varpi = \int_{4^{-1/3}}^{\infty} \frac{dx}{\sqrt{x^3 - 1/4}} = \frac{1}{2 \pi}\, \Gamma(1/3)^3 $$ (the zeros of $\wp'(z) - \sqrt{3}$ can be also determined). Since $$ \wp'(z)^2 - 3 = 4 \left[\wp(z)^3 - 1\right] $$ and $\wp(x)$ is real for $0 < x < \varpi$, we also get $$ \wp\left(\pm\frac{\varpi}{3}\right) = 1 $$ or, equivalently, $$ \int_1^{\infty} \frac{dx}{\sqrt{4x^3 - 1}} = \int_{\wp\left(\frac{\varpi}{3}\right)}^{\infty} \frac{dx}{\sqrt{4x^3 - 1}} = \frac{\varpi}{3} = \frac{1}{6 \pi}\, \Gamma(1/3)^3 $$ (the detailed calculations can be found in arXiv:2105.04307).

Since I am not an expert in elliptic functions, my question is whether the above facts are known.