= Solution
For the <period lattice> $\Lambda=\mathbb Z\omega_1+\mathbb Z\omega_2$, define the <Weierstrass elliptic function> by
$$
\wp(z)=\frac1{z^2}+\sum_{0\ne\omega\in\Lambda}\left(\frac1{(z-\omega)^2}-\frac1{\omega^2}\right).
$$
On a <compact set> away from $\Lambda$, the summand is $O(|\omega|^{-3})$ uniformly for sufficiently large $|\omega|$. The permitted lattice convergence estimate therefore gives <Normal convergence of the Weierstrass elliptic-function series>. Consequently $\wp$ is a <meromorphic function>, with a double <pole> of principal part $(z-\omega)^{-2}$ at each lattice point and no other <poles>. Termwise differentiation gives the normally and absolutely convergent series
$$
\wp'(z)=-2\sum_{\omega\in\Lambda}(z-\omega)^{-3}.
$$
Pairing $\omega$ with $-\omega$ shows that $\wp$ is even and $\wp'$ is odd. Reindexing the absolutely convergent derivative series gives $\wp'(z+\lambda)=\wp'(z)$ for every $\lambda\in\Lambda$. Thus $\wp(z+\lambda)-\wp(z)$ is constant, with removable singularities at the lattice points. For either basis period $\lambda=\omega_1,\omega_2$, evaluate at $z=-\lambda/2$: evenness makes this constant <zero>. These two <periods> imply all periods in $\Lambda$. Hence \b[$\wp$ is a nonconstant <elliptic function> with <period lattice> $\Lambda$.]
To obtain the <Laurent coefficients of the Weierstrass elliptic function>, take $0<|z|<\min_{\omega\ne0}|\omega|$ and expand
$$
\frac1{(z-\omega)^2}-\frac1{\omega^2}=\sum_{j\geq1}(j+1)\frac{z^j}{\omega^{j+2}}.
$$
The expansion and lattice summation can be interchanged uniformly on smaller disks, using the same convergent majorant. The odd powers cancel in pairs. In terms of the <lattice Eisenstein sums>, the answer is
$$
\boxed{\wp(z)=z^{-2}+\sum_{r\geq1}(2r+1)G_{2r+2}(\Lambda)z^{2r}
=z^{-2}+3G_4z^2+5G_6z^4+O(z^6).}
$$
Only the absolutely convergent <lattice Eisenstein sums> of weights at least four occur here; no conditionally defined $G_2$ is needed.
Put $g_2=60G_4$ and $g_3=140G_6$. The <Laurent series> just found gives
$$
\wp'(z)^2-4\wp(z)^3=-60G_4z^{-2}-140G_6+O(z^2).
$$
Therefore $F=\wp'^2-4\wp^3+g_2\wp+g_3$ has no <pole> at <zero>, and its value there is <zero>. By periodicity it has no <poles> anywhere. An entire <elliptic function> is bounded on a closed <fundamental parallelogram>, hence on the plane, and is constant by <Liouville theorem>. Since $F(0)=0$, this proves the <Weierstrass elliptic differential equation>
$$
\wp'^2=4\wp^3-g_2\wp-g_3.
$$
We still need to identify the three roots without assuming their distinctness. For a nonconstant <elliptic function> $u$, the <argument principle> applied to $u'/u$ around a <fundamental parallelogram> makes the opposite-edge integrals cancel. Thus its <zeros> and <poles>, counted with <multiplicity>, have equal total number. In particular $\wp-a$ has exactly two <zeros> in a period cell for every finite $a$.
Let $h_i=\omega_i/2$, with $\omega_3=-\omega_1-\omega_2$. Their classes are the three nonzero <two-torsion points of a complex torus>. Oddness and periodicity give $\wp'(h_i)=-\wp'(h_i)$, so $\wp'(h_i)=0$. Thus $\wp-e_i$, where $e_i=\wp(h_i)$, has a <zero> of <multiplicity> at least two at $h_i$, and the preceding count says its <multiplicity> is exactly two. If two of the <half-period values of the Weierstrass elliptic function> were equal, the same function would have at least four <zeros>, a contradiction. Evaluating the <Weierstrass elliptic differential equation> at the three half-periods now gives three distinct roots of its cubic. Consequently
$$
\boxed{\wp'(z)^2=4\prod_{i=1}^3(\wp(z)-e_i).}
$$
For the remaining identity, fix $i$ and put $A_i=(e_i-e_j)(e_i-e_l)$, where $\{i,j,l\}=\{1,2,3\}$. The <elliptic function>
$$
H_i(z)=(\wp(z+h_i)-e_i)(\wp(z)-e_i)
$$
has possible <poles> only at $z=0,-h_i$ modulo $\Lambda$. At each, the other factor has an exact double <zero>, so these <poles> are removable. Thus $H_i$ is constant, and its limit at <zero> is $\wp''(h_i)/2$. Differentiating the cubic <differential equation> first where $\wp'\ne0$, then extending the resulting identity holomorphically, gives $\wp''=6\wp^2-g_2/2$. Since $e_i$ is a root of $P(X)=X^3-g_2X/4-g_3/4$, we obtain
$$
\frac12\wp''(h_i)=3e_i^2-\frac{g_2}{4}=P'(e_i)=A_i,
\qquad
\wp(z+h_i)=e_i+\frac{A_i}{\wp(z)-e_i}.
$$
This proves the <Weierstrass half-period translation formula> by <pole> cancellation. Set $u_i=h_i/2=\omega_i/4$. Since $u_i+h_i\equiv-u_i\pmod\Lambda$, evenness gives $\wp(u_i+h_i)=\wp(u_i)$, and therefore
$$
A_i=(\wp(u_i)-e_i)^2.
$$
The denominator here is nonzero: the only <zero> of $\wp-e_i$ modulo $\Lambda$ is $h_i$, whereas $u_i$ is not congruent to $h_i$. Finally differentiate the <Weierstrass half-period translation formula> and use $z-h_i\equiv z+h_i\pmod\Lambda$ to obtain the <Weierstrass quarter-period derivative identity>
$$
\boxed{\frac{\wp'(z-\omega_i/2)}{\wp'(z)}=-\left(\frac{\wp(\omega_i/4)-e_i}{\wp(z)-e_i}\right)^2.}
$$
The equality holds wherever these quotients are initially defined, and hence as an identity of <meromorphic functions> at all remaining points.
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