Let $f(w_1,\ldots,w_n;z)$ be a holomorphic function of $n+1$ variables. For every fixed $w_1\ldots w_n$, let $g(w_1,\ldots,w_n;z)$ be an analytic continuation of $f$ as a holomorphic function of $z$. Of course, $g$ is holomorphic in $z$. Now, is $g$ holomorphic in $w_1,\ldots, w_n,z$ as well?
In the concrete, I am interested in the following situation. Let $E(a_1,\ldots,a_n;z)$ be an ODE, where $z$ is a complex variable and $a_1,\ldots,a_n$ are complex parameters. Assume that $E(a_1,\ldots,a_n;z)$ has a local solution $u(a_1,\ldots,a_n;z)$ which is holomorphic in $a_1,\ldots,a_n,z$, and $v(a_1,\ldots,a_n;z)$ is an analytic continuation of $u$. Is $v$ holomorphic in $a_1,\ldots,a_n,z$?
The answer is yes and it is a consequence of the classical Hartogs‘s theorem on separately holomorphic functions. Precisely, let $u(a_1,\dots,a_n,z)$ be a holomorphic function on a domain $D_u\in\mathbb{C}^{n+1}$ which is analytically continuable to a larger domain respect to the variable $z$ as a function $v(a_1,\dots,a_n,z)$: then
$v$ is separately holomorphic respect to $a_1,\dots,a_n$ for every $z\notin D_u$. To see this, choose $z_0\in D_u\cap\{z\in\mathbb{C}\}$ such that the Taylor series expansion of $v$ in $z_0$ has a convergence disk not entirely contained in $D_u\cap\{z\in\mathbb{C}\}$ and whose of radius $R_{z_0}\geq c>0$ does not depend on $a_1,\dots,a_n$. Such $z_0$ exists, since assuming the contrary would deny the possibility of analytically continue $u$ respect to $z$ outside $D_u\cap\{z\in\mathbb{C}\} $.
Then, evaluating this Taylor series at fixed
point $z_1\notin D_u\cap\{z\in\mathbb{C}\} $ inside its radius of convergence, we get
$$
\begin{split}
v(a_1,\dots,a_n, z_1)&=\sum_{k=0}^\infty\frac{1}{k!} \frac{\partial^k v}{\partial z^k}(a_1,\dots,a_n, z_0)(z_1-z_0)^k\\
&=\sum_{k=0}^\infty\frac{1}{k!} \frac{\partial^k u}{\partial z^k}(a_1,\dots,a_n, z_0)(z_1-z_0)^k
\end{split}\tag{1}\label{1}
$$
since $v=u$ on $D_u$. Now \eqref{1} implies that the $N$-th order Taylor polynomial
$$
v_N(\dots,a_j,\dots,z_1)=\sum_{k=0}^N\frac{1}{k!} \frac{\partial^k u}{\partial z^k}(\dots,a_j,\dots, z_0)(z_1-z_0)^k\tag{2}\label{2}
$$
can be considered as a sequence of holomorphic functions in each single variable $a_j$, converging uniformly (thanks to the holomophicity of $u$ and to the uniform convergence of \eqref{1}) to $v$. This suffices to prove the separate analyticity of $v$, analytic continuation of $u$, respect to all its variables in the domain $$ D_u\cup\{(a_1,\dots,a_n,z)|0<|z-z_0|\leq |z_1-z_0|\}. $$ The same argument can be repeated to cover all the domain $D_v$, as quckly sketched in the following picture: