Let $\Omega$ be an open set of $\mathbb{C}^n$ and let $f$ be analytic in $\Omega$. Assume $P\in\mathbb{C}[z_1,\ldots,z_n]$ is a polynomial whose irreducible factors are all of multiplicity one.
If $f$ vanishes on the zero set $Z(P)$ of $P$, how to justify that $f/P$ is analytic in $\Omega$ ?
I think you could argue stalk wise. Take a point $x \in \Omega$. By general theory, the ring of convergent power series at $x$, $\mathcal O_{\mathbb C^n, x}$, is a UFD. Since $P$ is square-free, and $f$ vanishes at $\{P =0\}$, $P$ is a factor of $f$ in $\mathcal O_{\mathbb C^n,x}$ by the analytic Nullstellensatz. So there is a quotient $\overline{g_x} = f/P$ in $\mathcal O_{\mathbb C^n,x}$. This is represented by an actual holomorphic function $g_x \in \mathcal O(U_x)$ in some neighbourhood $U_x$ of $x$, and $g_x \cdot P = f$ holds on all of $U_x$. By the identity theorem, the $g_x$ glue to a holomorphic function $g$ on $\Omega$.
Note however, that the ring $\mathcal O(\Omega)$ is in general not a UFD. For example if $n = 1$ and $\Omega = \mathbb C$, the function $f(z) = \sin(z)$ has infinitely many irreducible factors $P_k = (z - k \pi), \,k \in \mathbb Z$.