= Solution
By the <argument principle>, the number of zeros minus poles, counted with multiplicity, in the fundamental parallelogram is $(2\pi i)^{-1}\oint f'/f\,dz$. The logarithmic <derivative> is periodic with the same lattice, so the <integrals> over opposite sides cancel. Thus \b[the numbers of zeros and poles are equal].
The <Weierstrass elliptic function> gives a degree-two map $\wp:\mathbb C/\Lambda\to\mathbb P^1$ whose fibers identify $z$ and $-z$. An even <elliptic function> is constant on these fibers, so it descends to a single-valued function $Q$ on the <sphere>. This is meromorphic even at a branch value: near a half-period the involution is $t\mapsto-t$, the <Laurent series> of $f$ has only even powers, and $\wp-\wp_0$ has a double zero and provides a local coordinate in $t^2$. At the image of $0$, $\wp\sim t^{-2}$ gives the same conclusion at infinity. A <meromorphic function> on the <Riemann sphere> is a <rational function>, so $\boxed{f=Q(\wp)}$.
\b[A meromorphic antiderivative need not be elliptic.] Take $f=\wp$. Its double poles have zero <residue>, so integrating on the punctured plane gives a single-valued primitive, since every closed-contour <integral> is a sum of zero <residues>. The primitive extends meromorphically over each pole, with local principal part $-1/z$. If $F$ were elliptic, it would have precisely one pole modulo the lattice, with <residue> $-1$, contradicting the zero sum of <residues> in a fundamental parallelogram. This is a nonconstant counterexample satisfying the stated premise; the preceding descent proves that <even elliptic functions are rational in the Weierstrass function>.
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