= Solution
Uniqueness in the defining characterization gives $\wp(-z)=\wp(z)$: the reflected function has the same <poles>, periodicity and prescribed <principal part>. Hence $\wp'$ is odd and periodic. At each nonzero half-period $\omega\in\{1/2,\tau/2,(1+\tau)/2\}$, $-\omega\equiv\omega\pmod\Lambda$, so $\wp'(\omega)=-\wp'(\omega)=0$.
An <elliptic function> has equally many zeros and <poles>, counted with multiplicities, in a <fundamental parallelogram>. This standard property applies to $\wp'$, whose sole <pole> modulo the lattice has order three. The three distinct nonzero half-periods already supply three zeros, so these are all its zeros and each is simple. Thus
$$
\boxed{\{z:\wp'(z)=0\}=\bigcup_{\omega\in\{1/2,\tau/2,(1+\tau)/2\}}(\omega+\Lambda).}
$$
Evenness and the prescribed expansion give $\wp(z)=z^{-2}+a_2z^2+a_4z^4+O(z^6)$. There is no constant term because the defining difference tends to zero, and only even powers occur. Consequently
$$
\wp'(z)^2-4\wp(z)^3=-20a_2z^{-2}-28a_4+O(z^2).
$$
Adding $20a_2\wp$ cancels the last <pole>. The resulting <elliptic function> is <holomorphic> everywhere and therefore constant, by boundedness on a <fundamental parallelogram> and <Liouville theorem>. Its value at zero is $-28a_4$. Thus
$$
\wp'^2=4\wp^3-20a_2\wp-28a_4=Q(\wp).
$$
At the three half-periods $Q(\wp(\omega))=0$. Their values are distinct: if two such points had the same value $e$, then $\wp-e$ would have a zero of <multiplicity> at least two at each, because its <derivative> vanishes there. That gives at least four zeros, contrary to its sole double <pole> and the zero-pole counting property. Hence the cubic has three distinct roots and leading coefficient four, proving
$$
\boxed{Q(w)=4(w-\wp(1/2))(w-\wp(\tau/2))(w-\wp((1+\tau)/2)).}
$$
All properties specific to $\wp$ used here were derived from its defining characterization; only the general elliptic zero-pole count and <holomorphic> constancy were used as standard facts.
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