Let $F,G \in \mathbb{C}[x,y]$. (We do not know if $F$ and $G$ have a common zero or not).
Assume that there exists $H(x,y) \in \mathbb{C}[x,y]$ with $H(0,0)=0$, such that $x$ divides $H(Fy, Gy)$.
When (= under which additional conditions) $x$ divides both $F$ and $G$?
A special case: $H(x,y)=\lambda x+\mu y$. Then $H(Fy, Gy)= \lambda Fy +\mu Gy= y(\lambda F + \mu G)$. Then $x$ divides $\lambda F + \mu G$. However, it may happen that $x$ does not divide either $F$ or $G$, for example, $\lambda=\mu=1$, $F=x+1$, $G=x-1$.
If $x$ divides $F$, then $x$ also divides $G$, and vice versa.
Any hints and comments are welcome!
Fix the value of $y$ used in $F, G$, and just consider them functions of $x$ only, and include the $y$ multiples in the definitions of these functions instead of listing it separately. So $F = F(x), G = G(x)$ and we are given that $H(0,0) = 0$ and that $H(F(0), G(0)) = 0$, and want conditions so that this implies that $F(0) = 0$ and $G(0) = 0$.
This makes it obvious: the only such condition on $H$ alone is that $H$ can have only one root, namely the given one: $(0,0)$. If it has any other root, then we can choose $F, G$ so that $(F(0), G(0))$ is this root.
But this is impossible. By the fundamental theorem of algebra, for each $y$, either $H(x,y)$ is constant with respect to $x$, or else it has at least one root $x_y$, so $H(x_y, y) = 0$. But if $H$ is constant with respect to $x$ for all $y \ne 0$, then $H$ does not depend on $x$ at all, and therefore $H(F(0), G(0)) = 0$ would not imply that $F(0) = 0$.
So there is no condition on $H$ that will work. The only conditions on $F, G$ that could be added are that $(F(0), G(0))$ must miss the other roots of $H$.