Let $R$ be any integral domain, and $R[x,y]$ the ring of polynomials over $F$ in two variables. If we regard $R[x,y]$ as $\left(R[x]\right)[y]$, i.e. as polynomials in $y$ whose coefficients come from $R[x]$, then given any two polynomials $p(x,y), d(x,y)$, as long as $d$ is monic we can use the Euclidean algorithm to find $q_1(x,y)$ and $r_1(x,y)$ so that $$p(x,y) = d(x,y)q_1(x,y)+r_1(x,y)$$ with $\textrm{ydeg } r_1 < \textrm{ydeg } d$ (where here, $\textrm{ydeg }$ denotes the degree in $y$, e.g. $x^2y^3$ has $\textrm{ydeg } =3$. Likewise if we regard $R[x,y]$ as $\left(R[y]\right)[x]$, i.e. as polynomials in $x$ with coefficients in $R[y]$, then given the same two polynomials $p(x,y), d(x,y)$ we can use the Euclidean algorithm to find $q_2(x,y)$ and $r_2(x,y)$ such that $$p(x,y) = d(x,y)q_2(x,y)+r_2(x,y)$$ with $\textrm{xdeg } r_1 < \textrm{xdeg } d$.
This shows that the Euclidean algorithm can be used on $R[x,y]$ in two distinct ways. In general the two quotients $q_1, q_2$ and remainders $r_1, r_2$ are different depending on which of those two paths we choose. Essentially when we choose to measure the degree of a polynomial by focusing on one variable or the other, we "break" the symmetry between $x$ and $y$.
Is there a way to define a Euclidean degree function and implement the Euclidean algorithm on $F[x,y]$ that is "even-handed", in the sense that it deals with both variables in the same way? The obvious thing to try is to use the total degree, but that doesn't work: with $p(x,y) = x^2y$ and $d(x,y) = xy^2$ there is clearly no $q(x,y), r(x,y)$ such that $x^2y = xy^2q(x,y) + r(x,y)$ and for which $r$ has total degree $< 3$.
My instinct is that it can't be done, but I am wondering if there is a simple way to prove it.
I guess a sub-question here is: Is there a name for a ring (like $R[x]$ and $R[x,y]$ for $R$ a domain) in which you can always divide a given polynomial by an arbitrary monic divisor, but not necessarily by any divisor? It's not quite as strong as being a Euclidean domain, but it's precisely the situation I'm thinking about.