So I was studying some stuff a bout ring theory and ended up with the question in the title. This is what I have so far:
Suppose $\bar{x}$ is reducible, i.e. $\bar{x}=\bar{p}\bar{q}$, for some non trivial $\bar{p},\bar{q}\in A$, now \begin{eqnarray} \bar{x}=\bar{p}\bar{q} &\Longrightarrow& x=pq+t(x^2-y^3)\quad\text{for some }p,q,t\in B\\ &\Longrightarrow&1=deg(x)=deg(pq+t(x^2-y^3))\\ &\Longrightarrow°(pq)=deg(t(x^2-y^3))\\ &\Longrightarrow°(p)+deg(q)=3+deg(t) \end{eqnarray} but doesn't seem to get any better. I have also tried to apply this property: If $A$ is a domain and $(a)\triangleleft A$ is prime, then $a$ is irreducible. But the ideal $(\bar{x})$ doesn't seem to be prime so $A/(\bar{x})$ is not domain (unless there is something I'm missing). That's what I have, any hints or approaches will be very well received. Thanks a lot.
First define a special degree $d(f)=3deg_x(f)+2deg_y(f)$. Using this you can see that the only monomial in $pq$ with $d<6$ is $x$.
Consider that in some ring (like $\Bbb{R}[x^{1/6},y^{1/6}]$ you can specialize $y$ to be $x^{2/3}$. In that specialization you have $x=p(x,x^{2/3})q(x,x^{2/3})$. Thus, $deg_x(p(x,x^{2/3}))+deg_x(q(x,x^{2/3}))=1$. Using the previous paragraph, you can show that in this case the degrees must be integers, so, after moving a constant multiple from $p$ to $q$ we have $p(x,x^{2/3})=1$ and $q(x,x^{2/3})=x$ (or conversely). In order for this to be the case, you can show (by induction, that this is necessary to make higher degree terms cancel) we must have $p=1+r(x^2-y^3)$ and $q=x+s(x^2-y^3)$. But then $\bar{q}=\bar{x}$, so $x$ is irreducible.