**EDIT**

Let $\mathbb{C}^{m*}$ be the space of non zero polynomials of degree at most $d$ in two variables. So an element of this space is essentially $$ f:=f_{00} + f_{10} x + f_{01} y + \ldots f_{0d} y^d $$ where not all the coefficients are zero. Hence, this space can be identified with $\mathbb{C}^{m*}$, where $m = \frac{d(d+3)}{2}+1$.

Let $\psi: \mathbb{C}^{m*} \times \mathbb{C}^2 \rightarrow \mathbb{C} $ be the evaluation map, ie $$ \psi(f, x,y) = f(x,y) = f_{00} + f_{10} x + f_{01} y + \ldots f_{0d} y^d.$$ I have two questions:

1) Given $(f ,x_0,y_0) \in \mathbb{C}^{m*} \times \mathbb{C}^2$ such that $\psi(f, x_0,y_0) =0 $ and a polynomial $f_t$ that is sufficiently "close" to $f$, does there exist a point $x_t, y_t$ close to $x_0, y_0$ such that $$ \psi(f_t, x_t, y_t) =0 ~~?$$

Note that this would be true if $(x_0, y_0)$ was a smooth point of $f$. But I am not making this assumption.

2) Let $ \mathcal{S}$ be a dense subspace of $\mathbb{C}^{m*}$. Define the
space $\mathcal{F}$ as

$$ \mathcal{F} := \{ (f, x,y) \in \mathcal{S} \times \mathbb{C}^2: \psi(f, x,y) =0 \} $$

Is it true that
$$ \overline{\mathcal{F}} = \{ (f, x,y) \in \mathbb{C}^{m*} \times \mathbb{C}^2: \psi(f, x,y) =0 \} $$
where
$\overline{\mathcal{F}}$ denotes the closure of $\mathcal{F}$ inside
$\mathbb{C}^{m*} \times \mathbb{C}^2$?

The basic idea being that if $ \psi(f, x,y) =0$ but $f \notin \mathcal{S} $ we can choose a sequence $ f_n \in \mathcal{S} $ converging to $f$. And by 1), there exists a sequence (possibly non unique) $(x_n, y_n) \in \mathbb{C}^2$ such that $$ \psi(f_n, x_n, y_n) =0 $$ This sequence $(f_n, x_n,y_n)$ should converge to $(f, x,y)$.