As shown is Gerry Myerson’s answer, the answer is NO when $\mathbb F$ is countably infinite.
The answer is YES when $\mathbb F$ is uncountable, however.
Sketch of proof : since there are only countably many degrees, the polynomials will share a common degree on an uncountable set. This bound on the degree allows one to use interpolation, and to retrieve the whole of $f$.
More detailed proof : Denote by $d(x)$ the degree of the univariate polynomial $f(x,.)$ for $x\in {\mathbb F}$ (recall that the degree of the zero polynomial is $-\infty$), and put $U_d=\lbrace x \in {\mathbb F} | d(x)=d\rbrace$ for $d\in \lbrace -\infty \rbrace \cup {\mathbb N}$. Then the $U_d$ form a countable partition of $\mathbb F$, so at least one of the $U_d$, say $U_{n}$, is uncountable.
We may assume that $n>0$, as the cases $n=-\infty$ and $n=0$ are similar and simpler. Let $y_0,y_1, \ldots y_{n}$ be $n+1$ distinct values in $\mathbb F$, this is possible because $\mathbb F$ is uncountable. (if the characteristic of $\mathbb F$ is zero, we can simply take $y_i=i$). Using Lagrange interpolation, let us put
$L_k(y)=\frac{\prod_{j \neq k}{(x-x_j)}}{\prod_{j \neq k}{(x_k-x_j)}}$
for $0 \leq k \leq n$. Then one has, for any polynomial $P$ of degree $\leq n$ and any $y\in{\mathbb F}$,
$ P(y)=P(y_0)L_0(y)+P(y_1)L_1(y)+ \ldots +P(y_n)L_n(y) $
In particular, one has for any $(x,y)\in U_n \times {\mathbb F}$,
$ (1) \ f(x,y)=f(x,y_0)L_0(y)+f(x,y_1)L_1(y)+f(x,y_2)L_2(y)+ \ldots +f(x,y_n)L_n(y) $ The right-hand side is a fixed bivariate polynomial, let us denote it by $Q(x,y)$. Let $y\in {\mathbb F}$. Then the two univariate polynomials $f(.,y)$ and $Q(.,y)$ coincide on the uncountable set $U_n$, so they must coincide everywhere. Finally $f=Q$ everywhere and we are done.