This question is from Silverman's 'the arithmetic of elliptic curves', p167. Given a meromorphic function $f$ on $\mathbb{C}/\Lambda$, there is indeed a standard proof that $f \in \mathbb{C}(\wp,\wp')$. It goes something like this. Write $f$ as a sum of an even function and an odd function: $$ f(z) = \frac{f(z) + f(-z)}{2} + \frac{f(z) - f(-z)}{2} $$ Using this trick we may assume that $f$ is odd, or that $f$ is even. In fact we can assume $f$ is an even function, since if $f$ is an odd elliptic function then $\wp' \cdot f$ is an even elliptic function. Thus it is enough to show that if $f$ is an even elliptic function then $f \in \mathbb{C}(\wp)$.
For even elliptic functions $f$, the identity $$ \operatorname{ord}_w f = \operatorname{ord}_{-w} f $$ holds for all $w \in \mathbb{C}$. Furthermore, if $2 w \in \Lambda$, then $\operatorname{ord}_w f$ is even, because the $i$-th derivative satisfies $$ f^{(i)}(-w) = f^{(i)}(w) = (-1)^i f^{(i)}(-w) $$ for all odd values of $i$ (the first equality follows because $2 w \in \Lambda$, and the last equality is achieved by repeatedly applying the chain rule). Therefore $$ \operatorname{div}(f) = \sum_{w \in H} n_w ((w) + (-w))・・・① $$ for some set of integers $n_w$, where $H$ is half of a fundamental parallelogram for $\Lambda$, and the sum has only finitely many nonzero terms.
Question. How can we deduce ① from '$\operatorname{ord}_w f$ is even' ? Isn't $n_w$ and $\operatorname{ord}_w f$ equal ? I don't see where the condition '$\operatorname{ord}_w f$' was used.
$\wp'(w+z)$ is odd for all $2w=0$ so that $\wp'(w)=0$, since $\wp'$ has only one triple pole at $0$ it means that it has at most 3 zeros and hence $$div(\wp')=-3 [0]+\sum_{2w=0}[w]$$
Whence for $2s\not \in \Lambda$ you get that $$div(\wp-\wp(s))=-2[0]+[s]+[-s]$$
This latter formula stays true when $2s=0$.
If $f$ is meromorphic $\Lambda$ periodic then $g(z)=f(z)+f(-z)$ has $N$ poles counted with multiplicity (on $\Bbb{C/\Lambda}$), one of them at $b_1$ of order $n_1$ and first Laurent coefficient $c_1$, then $g-\frac{c_j}{(\wp-\wp(b_j))^n}$ has $N-2$ poles (if $b_1=0$ look instead at $g-c_j \wp^{n/2}$). Repeating you'll get that $g-\sum_i C_i \wp^i -\sum_j \frac{c_j}{(\wp-\wp(b_j))^{n_j}}$ has no poles: it is constant by the maximum modulus principle.
Doing the same with $\wp'(z)(f(z)-f(-z))$ you'll get that $f\in \Bbb{C}(\wp,\wp')$.