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 $\sum \lambda_i f_i(z) \neq 0$, $valuation( \sum \lambda_i f_i(z)) \leq N$.

Let $h_{k}(z)=\sum_{j=1}^b u_j(z)v_j^{(k)}(0)$.
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, $C^i_{d+i}/C^i_i \times \lambda_0 \times$ 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.

My answer is not complete.