This question asks for sufficient criteria for a polynomial in two variables over a field, to be irreducible.
Let $u= Ax^3+Bx^2y+Cxy^2+Dy^3+ax^2+bxy+cy^2+dx+ey+f \in k[x,y]$, for some field $k$ (further assume that $k=\mathbb{C}$, if this helps).
Assume that $u$ is irreducible; is it possible to find all the forms of such $u$?
I only know how to show that, after a change of variables, we can obtain $B=0$; this is true in general, regardless of irreducibility (a change of variables does not change irreducibility; in other words, an automorphism of $k[x,y]$ takes an irreducible polynomial to an irreducible polynomial).
A partial answer: It follows from one of the answers to the above mentioned question that $s(x)-t(y)$ is irreducible, if the degrees of $s$ and $t$ are coprime.
Thank you very much!
I think it should be the other way around: you can determine all the reducible $u$ by looking at the forms of possible factors. Any $u$ that does not fall into one of those cases is irreducible.
EDIT: To simplify slightly, let's take $A = 1$ and $B=0$ (in a complete solution you'll also have to consider the case $A=0$, but I won't). A reducible $u$ of degree $3$ must have a factor of degree $1$, let's say $x + p y + q$ (again I'm neglecting the case of factors that depend only on $y$...). The remainder of $u$ on division by $x + p y + q$ (for the variable $x$) is $$ \left( -{p}^{3}-Cp+D \right) {y}^{3}+ \left( a{p}^{2}- 3\,{p}^{2}q-Cq-bp+c \right) {y}^{2}+ \left( 2\,apq-3\,p{q}^{2 }-bq-dp+e \right) y+a{q}^{2}-{q}^{3}-dq+f $$ which must be identically $0$, i.e. the coefficient of each power of $y$ here must be $0$. This gives us four polynomials in $p, q$ and the coefficients of $u$. In order for $u$ to have a factor of this form, these polynomials must have a common zero.
Thus $u$ is reducible if there are $p$ and $q$ such that $$ \eqalign{D &= p^3 + C p\cr c &= -a p^2 + 3 p^2 q + C q + b p\cr e & = -2apq + 3 pq^2 + bq + dp\cr f & = - a q^2 + q^3 + d q\cr} $$
It's possible to eliminate $p$ and $q$ from those polynomials, obtaining polynomials in $C, \ldots, f$, however the results are rather horrendously complicated.
To be somewhat less ambitious, let's fix some of the coefficients, say $a=-2,b=-1,c=0,d=1,e=2,f=3$, and ask for what $C$ and $D$ we get a reducible $u = x^3 + C x y^2+D y^3-2 x^2-x y+x+2 y+3$. The equations become $$ \eqalign{D &= p^3 + C p\cr 0 &= 2 p^2 + 3 p^2 q + C q - p\cr 2 & = 4pq + 3 pq^2 - q + p\cr 3 & = 2 q^2 + q^3 + q\cr} $$ Using a Groebner basis in Maple, this is equivalent to $$ \eqalign{44217 D^3+ & 6069 D^2+86 D+3 = 0\cr C &= -{\frac {867\,{D}^{2}}{13}}-{\frac {68\,D}{13}}-\frac{1}{13}\cr q&=-{\frac { 88434\,{D}^{2}}{325}}-{\frac {18207\,D}{325}}-{\frac{609}{325}}\cr p&={ \frac {14739\,{D}^{2}}{325}}+{\frac {272\,D}{325}}-\frac{61}{325}}$$