If $e^{zw}=\frac{\sum_{i=1}^{a} f_i(z)g_i(w)}{\sum_{j=1}^b u_j(z)v_j(w)}$,
$e^{zw}(\sum u_j(z)v_j(w))=\sum f_i(z)g_i(w)$
$\frac{\partial^{n}}{\partial w^n}e^{zw}(\sum u_j(z)v_j(w))=\frac{\partial^{n}}{\partial w^n}\sum f_i(z)g_i(w)$
$\sum_{k=0}^n C^k_n z^k e^{zw}(\sum u_j(z)v^{(n-k)}_j(w))=\sum f_i(z)g_i^{(n)}(w)$
If $w=0$, $\sum_{k=0}^n C^k_n z^k(\sum u_j(z)v^{(n-k)}_j(0))=\sum f_i(z)g_i^{(n)}(0)$
For $(f_i(z))$ a finite family. $N \in \mathbb{N}$ exists such that, for all $(\lambda_i)$,if $\sum \lambda_i f_i(z) \neq 0$, $valuation( \sum \lambda_i f_i(z)) \leq N$.
Soit $h_{k}(z)=\sum_{j=1}^b u_j(z)v_j^{(k)}(0)$. The infinite family $(h_0,...,h_k,...)$ has rank less or equal than $b$.
We have:
$$h_0 \in Vect(f_i)$$ $$zh_0+h_1 \in Vect(f_i)$$ $$z^2h_0+2zh_1 +h_2\in Vect(f_i)$$ $$z^3h_0+3z^2h_1+3zh_2+h_3 \in Vect(f_i)$$
If $h_0=0$, a diagonal is empty.
If $h_1= \lambda h_0$, we can substract $\lambda \times$ the first line to the second line. We can substract $2\lambda \times$ the second line to the third. And substract $k\lambda \times $ the line $k$ to the line $k+1$, etc... A diagonal vanishes.
We can suppose that the family $(h_1,...,h_c)$ generates the infinite family $(h_0,...,h_k,...)$
If $h_d=\lambda_0 h_0 + ... \lambda_c h_c$, we substract to the $d+i+1$-th line, $C^i_{d+i}/C^0_i \times \lambda_0 \times$ line $1$-th, ...,$C^i_{d+i}/C^i_{j+i} \times \lambda_j \times$ line $(j+1)$-th...,$C^i_{d+i}/C^i_{c+i}\times \lambda_c \times$ line $(c+1)$-th.
The diagonal $d+1$ vanishes.
So, if $c \neq 0$, the valuation is not bounded in $Vect(f_i)$. Contradiction.
So, $c=0$. And,$h_i=0$ for all $i$. Hence $\sum f_i(z)g_i^{(n)}(0)=0$ for all $n$. So $e^{zw}=0$. Contradiction.