Let $\wp$ be the Weierstrass ellptic function. One can show that $\wp''(z) = 6\wp(z)^2 - \frac{g_2}{2}$ where $g_2 = \sum_{0\neq \omega \in L} \frac{60}{\omega^4}$. Since I want to investigate the number of zeroes of $\wp''$, knowing that $\wp$ has degree 2 (when looked as a map from the usual lattice $L = \{m\omega_1 + n\omega_2\}$ to $\mathbb{C}\cup \infty$) it really depends on whether $g_2 = 0$ or not.
So can we describe the pairs $(\omega_1, \omega_2)$ for which $g_2=0$? I am not even sure whether there are such pairs. I have no idea how to approach this.
$j$ has exactly one zero $\tau=e^{2i\pi /3}$ in the fundamental domain $\{ |z| \ge 1,\Im(z) >0,\Re(z)\in [-1/2,1/2)\}$ of $SL_2(\Bbb{Z}) \setminus \mathcal{H}$ so $j(z) =0$ iff $z = \frac{a\tau+b}{c\tau+d}$ for some $a,b,c,d\in \Bbb{Z},ad-bc=1$ and $g_2( u\Bbb{Z}+v\Bbb{Z})=0$ iff $(u,v) = (s(a\tau+b),s(c\tau+d))$ for some $a,b,c,d\in \Bbb{Z},ad-bc=1,s \in \Bbb{C}^*$
For any $L$, $\wp_L''(z) = 6\wp_L(z)^2 - \frac{g_2(L)}{2}$ has $4$ zeros counted with multiplicity, if $g_2(L) = 0$ or $6\wp_L(\omega/2)^2- \frac{g_2(L)}{2}=0$ it is $2$ double zeros, otherwise it is $4$ simple zeros