Let us prove the desired result for $n=2$. We have
$$f(x,y)=\frac{\sum_{i=0}^m a_i(y)x^i}{\sum_{i=0}^k b_i(y)x^i}=r_x(y),\tag{1}$$
where the $a_i$'s and $b_i$'s are some functions and, for each $x$, $r_x$ is a rational function. We want to show that $f$ is a rational function. Without loss of generality (wlog), $b_0=1$. It is then enough to verify the claim that the $a_i$'s and $b_i$'s are rational functions.

Let us do this by induction on the (total) degree $d:=m+k$ of $f$ in $x$. If $d=0$, then the claim is obvious. Suppose now that $d=m+k\ge1$. Wlog, $m\ge k$ (or take the reciprocal of $f$). Let
$$g(x,y):=\frac{r_x(y)-r_0(y)}x=\frac{f(x,y)-r_0(y)}x
=\frac{\sum_{j=0}^{m-1}c_j(y)x^j}{\sum_{i=0}^k b_i(y)x^i},$$
where $c_j(y):=a_{j+1}(y)-b_{j+1}(y)r_0(y)$, with $b_i(y):=0$ for $i>k$. Then $g(x,y)$ is of degree $<d=m+k$ in $x$ and is rational in $x$ and in $y$. So, by induction, all the $b_i$'s and all the $c_j$'s are rational functions, and hence all the $a_i$'s are rational functions, just as claimed.

As Wojowu noted, the above argument tacitly assumes that $m=m_y$ and $k=k_y$ do not depend on $y$. Also, (for uncountable fields) Wojowu showed how to fix this argument. His reasoning can now be used to prove the desired result for any $n\ge2$. This can be done by induction on $n$, as sketched below.

As noted by the OP, the function $f$ is meromorphic and hence defined on a nonempty open subset $E$ of $\mathbb C^n$. Let $x:=z_1$ and $y:=z_2,\dots,z_n$. For natural $d$, let $S_d:=\{y\colon\exists(x,y)\in E, D_y(f)\le d\}$, where $D_y(f)=m_y+k_y$ and $m_y=m,k_y=k$ with $m,k$ as in (1). The sets $S_d$ are closed in the open set $U:=\{y\colon\exists(x,y)\in E\}\subseteq\mathbb C^{n-1}$ and $\bigcup_d S_d=U$. So, by the Baire category theorem, for some natural $p$ the set $S_p$ contains a nonempty open ball $B$. Fixing now $z_3,\dots,z_n$ and using the above argument, we see that, for each $i$, $a_i(y)=a_i(z_2,z_3,\dots,z_n)$ is rational in $z_2$ (that is, in $z_2\in\{t\colon (t,z_3,\dots,z_n)\in B\}$). Similarly, $a_i(y)=a_i(z_2,\dots,z_n)$ is rational in $z_j$ for each $j\in\{2,\dots,n\}$. So, by induction on $n$, $a_i(y)=a_i(z_2,\dots,z_n)$ is rational in $z_2,\dots,z_n$, for each $i$. Similarly, $b_i(y)=b_i(z_2,\dots,z_n)$ is rational in $z_2,\dots,z_n$, for each $i$. Thus, $f$ is rational.