= Solution
The local estimate gives $\wp(z)=z^{-2}+O(z)$ at zero and, by periodicity, at every lattice point. If two candidates existed, their difference would extend to an entire doubly periodic function: the principal parts cancel at every possible pole. It is bounded on a fundamental parallelogram, hence on the plane, so <Liouville theorem> makes it constant. The local $O(z)$ difference tends to zero, making the constant zero. Thus \b[the function is unique].
The function $z\mapsto\wp(-z)$ has the same periodicity, poles and local estimate. Uniqueness therefore proves \b[$\wp(-z)=\wp(z)$]. Its Laurent expansion is consequently
$$
\wp(z)=z^{-2}+c_2z^2+c_4z^4+\cdots,
$$
with no constant term by the given estimate. Periodicity and evenness imply $\wp(1/2+t)=\wp(1/2-t)$, so $\wp'(1/2)=0$. The elliptic zero-pole counting theorem says a nonconstant <elliptic function> has, in one period cell, equally many zeros and poles counted with multiplicity. Applied to $\wp(z)-\wp(1/2)$, whose total pole order is two, it gives precisely two zeros counted with multiplicity. The zero at $1/2$ already has multiplicity at least two. It therefore has exactly two, proving
$$
\boxed{\wp''(1/2)\ne0.}
$$
Finally $\wp''-6\wp^2$ is periodic, and the Laurent expansion shows cancellation of all possible pole terms: $\wp''=6z^{-4}+2c_2+O(z^2)$ and $6\wp^2=6z^{-4}+12c_2+O(z^2)$. The difference extends to an entire bounded function and is constant. Hence
$$
\boxed{\wp''(z)=6\wp(z)^2+A,\qquad A=-10c_2.}
$$
Every property of the <Weierstrass elliptic function> used here has been obtained from the stated local and periodic properties, together with the explicitly stated standard elliptic-function counting theorem.
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