This is an exercise adapted from Apostol. The problem is stated as
Prove that $$\wp''\left(\frac{\omega_1}{2}\right)=2(e_1-e_2)(e_1-e_3)$$ where $\omega_1,\omega_2$ generates the lattice for $\wp$.
I could see that by Weierstrass' differential equation, we have $$2\wp''\wp'=4\wp'((\wp-e_1)(\wp-e_2)+(\wp-e_2)(\wp-e_3)+(\wp-e_3)(\wp-e_1))$$ and $$2\wp'''\wp'+2\wp''^2=4\wp''(\cdots)+4\wp'(\cdots)$$ At $z=\frac{\omega_1}{2}$ we have $\wp'''=\wp'=0$ since they are odd elliptic functions. Therefore, $$2\wp''^2\left(\frac{\omega_1}{2}\right)=4(e_1-e_2)(e_1-e_3)\wp''$$
Now if $\wp''\left(\frac{\omega_1}{2}\right)\neq0$ we are done. However, I find it very difficult to prove the claim. I tried the following steps: first, we assume $\wp'(z)\neq 0$. Then by the expanded differential equation, $$\wp''=6\wp^2-\frac{1}{2}g_2=6\left(\wp-\sqrt{\frac{g_2}{12}}\right)\left(\wp+\sqrt{\frac{g_2}{12}}\right)$$ Since $\wp(z)\pm\sqrt{\frac{g_2}{12}}$ are double zeros of $\wp''$ and the order of $\wp''$ is $4$, if I could prove that $\pm\sqrt{\frac{g_2}{12}}\neq e_i$, $i=1,2,3$ then we are done.
I tried to use the fact that $g_2=2(e_1^2+e_2^2+e_3^2)$, but it left me with the problem of proving $$\frac{e_1^2+e_2^2+e_3^2}{6}\neq e_i^2$$ for all $i$.
How can I proceed? or is there a simpler argument that applies? I must have missed something. Please help me. Thanks in advance.
A very simple argument is counting the multiplicities of values. $\wp'$ is an odd elliptic function with a single pole of order $3$, hence it attains each value in $\widehat{\mathbb{C}} = \mathbb{C}\cup \{\infty\}$ exactly three times (in a fundamental region) counting multiplicities. Now by oddness, we have $$\wp'\biggl(\frac{\omega_1}{2}\biggr) = \wp'\biggl(\frac{\omega_2}{2}\biggr) = \wp'\biggl(\frac{\omega_1 + \omega_2}{2}\biggr) = 0\,,$$ hence these are all simple zeros of $\wp'$. In other words, $\wp''$ doesn't vanish at the half-periods.