= Solution
Write $D(f)=\mathbb C\setminus f^{-1}(0)$. The zero and constant-one polynomials give $D(0)=\varnothing$ and $D(1)=\mathbb C$. Also,
$$
D(f)\cap D(g)=D(fg),
$$
so $\tau$ is closed under finite intersections. Every nonzero one-variable complex polynomial has only finitely many roots, and every finite subset $\{a_1,\ldots,a_m\}$ is the zero set of $\prod_j(z-a_j)$. Consequently $\tau$ is exactly the <cofinite topology>: its open sets are the empty set and the complements of finite sets. An arbitrary union of such sets is again empty or has finite complement, so $\tau$ is a <topology>.
The <product topology> on $X\times Y$ is the topology with basis
$$
\{U\times V:U\subseteq X\text{ open},\ V\subseteq Y\text{ open}\}.
$$
The proposed complement need not be open. Take
$$
g(z,w)=z-w.
$$
Then $g^{-1}(0)$ is the diagonal. If its complement were open, a point such as $(0,1)$ would have a basic neighbourhood $U\times V$ contained in that complement. But $U$ and $V$ are nonempty cofinite subsets of $\mathbb C$, so $U\cap V\ne\varnothing$. For $t\in U\cap V$, the point $(t,t)$ belongs both to $U\times V$ and to the diagonal, a contradiction. Hence
$$
\boxed{\mathbb C^2\setminus g^{-1}(0)\text{ is not always product-open}}.
$$
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