= Solution
Write the <period lattice> as $\Lambda=\mathbb Z\omega_1+\mathbb Z\omega_2$, with $\omega_1,\omega_2$ real-linearly independent and $\operatorname{Im}(\omega_2/\omega_1)>0$. A <meromorphic function> $f$ is an <elliptic function> for this lattice if $f(z+\lambda)=f(z)$ for every $\lambda\in\Lambda$. If it has no <poles>, it is an <entire function> bounded on the closure of a <fundamental parallelogram>. Periodicity makes it bounded on all of $\mathbb C$, so the <Liouville theorem> proves that \b[an elliptic function without poles is constant].
Here are the two contour identities needed for counting its zeros and their sum. Choose a positively oriented <fundamental parallelogram> $P$ whose boundary avoids all zeros and <poles>, and put $h=f'/f$. The <argument principle> gives
$$
N-P_f=\frac1{2\pi i}\int_{\partial P}h(z)\,dz=0,
$$
because $h$ is periodic and opposite edges cancel. Here $N,P_f$ count zeros and <poles> with multiplicity. For the weighted integral, let $J_i$ be the integral of $h$ along the edge from a vertex $c$ to $c+\omega_i$. Translating the opposite edges gives
$$
\frac1{2\pi i}\int_{\partial P}z h(z)\,dz
=\frac{\omega_1J_2-\omega_2J_1}{2\pi i}.
$$
The endpoint values of $f$ agree along either edge. A continuous logarithm along the edge therefore changes by $2\pi i n_i$, with $n_i\in\mathbb Z$, and $J_i=2\pi i n_i$. On the other hand, the <residue theorem> evaluates the left side as the sum of the zero positions minus the sum of the pole positions, both with multiplicity. Thus the <elliptic divisor-sum identity> is
$$
\sum_j r_jz_j-\sum_l m_lp_l=\omega_1n_2-\omega_2n_1\in\Lambda.
$$
Under the stated pole hypothesis there is just one pole class, represented by zero, of order $m$. Consequently
$$
\boxed{N=m,\qquad\sum_j r_jz_j\equiv0\pmod\Lambda.}
$$
Changing representatives of the zero classes changes their weighted sum by an element of $\Lambda$, so the congruence is well defined.
The <Weierstrass elliptic function> is
$$
\wp_\Lambda(z)=\frac1{z^2}+\sum_{0\ne\omega\in\Lambda}
\left(\frac1{(z-\omega)^2}-\frac1{\omega^2}\right).
$$
On a bounded set of $z$, the summand is $O(|\omega|^{-3})$ for large $|\omega|$. The lattice sum of $|\omega|^{-3}$ converges in real dimension two, proving <Normal convergence of the Weierstrass elliptic-function series> on compact sets away from $\Lambda$. Hence $\wp$ is a <holomorphic function> there, and at zero its principal part is $z^{-2}$, so it has a double pole. Replacing $\omega$ by $-\omega$ shows that $\wp$ is even. Differentiating the normally convergent series gives
$$
\wp'(z)=-2\sum_{\omega\in\Lambda}(z-\omega)^{-3},
$$
whose absolute convergence permits reindexing by any $\lambda\in\Lambda$. Thus $\wp'$ is periodic and $\wp(z+\lambda)-\wp(z)$ is constant. Substituting $-z-\lambda$ for $z$ and using evenness changes that constant to its negative, so it is zero. This proves that $\wp$ is an <elliptic function>, with double <poles> precisely at the lattice points.
The preceding zero count applied to $\wp-a$ says that it has exactly two zeros modulo $\Lambda$, counted with multiplicity. Evenness pairs any zero $z$ with $-z$. If these are distinct classes they must both be simple. If they coincide, $2z\in\Lambda$, and locally $\wp(z+t)=\wp(z-t)$: the zero has even order, which the total count forces to be two. In particular such a zero cannot be a lattice point, where $\wp$ has a pole. \b[Every finite fibre of the Weierstrass function consists of two opposite simple points or one double nonzero half-period point.]
To obtain the rational representation, first consider an even <elliptic function> $g$. The preceding description proves that $\wp:\mathbb C/\Lambda\to\mathbb P^1$ has exactly the fibres $\{z,-z\}$, so $g$ defines a function $A$ of $w=\wp(z)$. Away from the branch values a local inverse of $\wp$ makes $A$ <meromorphic>. At a half-period $z_0$, the local expansion is $w-w_0=c t^2+O(t^4)$, with $c\ne0$ because the zero has order exactly two. The Laurent series of $g(z_0+t)$ contains only even powers, and $t^2$ is a holomorphic local coordinate as a function of $w-w_0$. Thus $A$ is <meromorphic> at that value too. At zero the same argument uses $1/w=t^2+O(t^6)$, proving meromorphy at infinity. A <meromorphic function> on the <Riemann sphere> is a <rational function>: subtract its finitely many principal parts and use compactness to make the remainder constant. This proves that <even elliptic functions are rational in the Weierstrass function>.
For an arbitrary <elliptic function>, split $f=f_++f_-$, where $f_\pm(z)=(f(z)\pm f(-z))/2$. The even part is $A(\wp)$. Since $\wp'$ is odd and not identically zero, $f_-/\wp'$ is an even <meromorphic> <elliptic function>, including at the zeros of $\wp'$ where the quotient may have <poles>. It is therefore $B(\wp)$ for a <rational function> $B$. The <elliptic function-field decomposition> is
$$
\boxed{f(z)=A(\wp(z))+B(\wp(z))\wp'(z),\qquad A,B\in\mathbb C(w).}
$$
The representation is unique: its even and odd parts determine $A$ and $B$, and $\wp$ takes every value on the <Riemann sphere>.
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