= Solution
Put $R(z)=\tfrac14(\wp''/\wp')^2-2\wp(z)$. This is an <elliptic function>. Its only possible poles modulo $\Lambda$ are $0$ and the three simple zeros $h$ of $\wp'$. At such a nonzero half-period, periodicity and evenness give $\wp(h+t)=\wp(h-t)$, so
$$
\wp'(h+t)=At+O(t^3),\quad A\ne0,
\qquad \frac{\wp''(h+t)}{\wp'(h+t)}=\frac1t+O(t).
$$
It follows that $R(h+t)=1/(4t^2)+O(1)$ with no simple-pole term. This is exactly the principal part of $\wp(2h+2t)=\wp(2t)$.
At zero the source expansion yields
$$
\wp=z^{-2}+O(z^2),\quad\wp'=-2z^{-3}+O(z),\quad
\wp''=6z^{-4}+O(1),\quad\wp''/\wp'=-3z^{-1}+O(z^3).
$$
Consequently $R(z)=1/(4z^2)+O(z^2)$, again matching $\wp(2z)$. Thus $R(z)-\wp(2z)$ is a <holomorphic function> on a compact connected <Riemann surface>, hence constant. Its limit at zero is zero, proving
$$
\boxed{\wp(2z)=\frac14\left(\frac{\wp''(z)}{\wp'(z)}\right)^2-2\wp(z).}
$$
The equality extends over the apparent singularities as an identity of <meromorphic functions>.
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