Let $f \in k[x,y]$, $k$ is a field of characteristic zero. Assume that $(a,b) \in k^2$ is a root of $f$, namely, $f((a,b))=0$.
Is it true that $(x-a)(y-b)$ divides $f$, namely, $f=(x-a)(y-b)g$, for some $g \in k[x,y]$?
Of course, in case of one variable, if $a \in k$ is a root of $f \in k[x]$, then $f=(x-a)g$, see this.
Remarks: (1) The answers of this question are perhaps relevant. (2) Perhaps we should consider the given $f \in k[x,y]$ as a polynomial in one variable $x$ over the integral doamin $k[y]$ and also as a polynomial in one variable $y$ over $k[x]$.
Edit: After receiving a comment with a counterexample, what if we further assume that $(a,b)$ is also a root of $f_x$ and a root of $f_y$; is now $(x-a)(y-b)$ necessarily divides $f$?
Perhaps the answer is positive by considering $f \in k[y][x]$ and $f \in k[x][y]$, though I am not sure, since it is just one point $(a,b)$, not a line. An example I had in mind is: $f=xy=(x-0)(y-0)$, $(a,b)=(0,0)$, $f_x=y$, $f_y=x$. Notice that also $(x,0)$ and $(0,y)$ are roots of $f$, so regardless of the derivatives, $(x,0)$ being a root of $f$ implies that $y-0$ divides $f$, and $(0,y)$ being a root of $f$ implies that $x-0$ divides $f$. Perhaps the resultant will help?
The answer to this question seems relevant, especially: "if $p=p(X_1,X_2,\ldots,X_n)$ is a polynomial over an algebraically closed field $F$ and $q$ is a nonzero polynomial over $F$ in $n$ variables which is irreducible, and such that $q(X)=0\implies p(X)=0$, then $q$ divides $p$. This follows from Hilbert's Nullstellensatz", which is probably very close to what I was looking for.
