There is also a more algebraic perspective.

For this, it is easier (at least for me) to ``reverse engineer'' things by starting with your points, and then deducing something about the elliptic function values.

I will working on the elliptic curve $E: y^2 = 4 x^3 - 1$, which is identified with the quotient $\mathbb C/\mathbb T$ via $z \bmod \mathbb T \mapsto \bigl( \wp(z), \wp'(z) \bigr).$

Let $P$ be the point of $E$ which is the image of $\varpi/3$. It is a point of order $3$ in $E$. Now $E$ has ``complex multiplication'' by the ring $\mathcal O$ obtained by adjoining $\zeta = e^{2\pi i/3}$ to $\mathbb Z$ (because the lattice $\mathbb T$ is a submodule of $\mathbb C$ with respect to this ring). The element $\zeta$ sends $(x,y)$ to $(\zeta x, y)$.

In the ring $\mathcal O$, the number $3$ is not prime: it factors as
$3 = (1 - \zeta)(1 - \zeta^{-1}) = (1-\zeta)^2 \cdot (-\zeta^{-1}).$

So among the $3$-torsion points, there are some even more special points, namely those killed by multiplication by $(1-\zeta)$ (and not just by $(1-\zeta)^2$). Looking at the formula for multiplication by $\zeta,$
we see these are the points with $x = 0$, so the points $(0,\pm i).$

There are $9$ points of order dividing $3$ altogether, namely the origin (i.e. the point at infinity), these $2$ points $(0,\pm i)$, and then the remaining $6$ points $(x,y)$ which are killed by $3$, but not by $(1-\zeta)$.

If $P$ is one of these latter points, then we see that $[1-\zeta] P = (0,\pm i).$

We can compute the $x$-coordinate of $[1-\zeta] P$ explicitly, in terms of the $x$-coordinate of $P$, just using the usual addition formula. The
answer is
$$x^3 - 1 = 0.$$

So we find that these $6$ points satisfy $x^3 = 1$, and then (from the equation for $E$) $y^2 = 3$.

In terms of the lattice, these points are the cosets of $\varpi/3,$
$2 \varpi/3$, $\zeta \varpi/3$, $\zeta 2 \varpi/3$, $\zeta^{-1}\varpi/3,$ and $\zeta^{-1} 2 \varpi/3$. (The remaining cosets of order $3$ are exactly
$0,$ $(1 + \zeta) \varpi/3$, and $2(1+\zeta)\varpi/3$, and when you multiply these by $(1-\zeta)$, you actually land in $\mathbb T$.)

So there are $6$ lattice cosets, and $6$ sets of $(x,y)$-coordinates, which match when you apply $z \bmod \mathbb T \mapsto \bigl(\wp(z),\wp'(z)\bigr).$

The cosets $\varpi/3$, $\zeta \varpi/3$, and $\zeta^{-1} \varpi/3$ are related by successive multiplications by $\zeta,$ and so have the same $y$-value. (Either $\sqrt{3}$ or $-\sqrt{3}$.)

Similarly the cosets $2\varpi/3$, $\zeta 2 \varpi/3,$ and $\zeta^{-1} 2 \varpi/3$ have the same $y$-value. (Again, either $\sqrt{3}$ or $-\sqrt{3}$.)

To see that $\varpi/3$ actually maps to $-\sqrt{3}$, it it just a matter of checking that $\wp'(\varpi/3)$ is negative, which I'd guess one can fairly easily do.

This answer may look elaborate, but I've tried to include details to help non-experts in this perspective. I think it's fair to say that typically all such identities are part of the general theory of complex multiplication, and derivable by this kind of analysis.