As the title says is $$x^3+y^2x^2+3yx-y$$ irreducible in $\mathbb{Q}[y][x]$ ? Apparently the answer is that it is indeed irreducible.
My attempt:
My first thoughts are what is $\mathbb{Q}[y][x]$, to which I reckon it's something like $(\mathbb{Q}[y])[x]$, i.e. polynomials in $x$ with coefficients in $\mathbb{Q}[y]$, polynomials in $y$.
Assuming this is correct, then maybe I can use Gauss' Lemma? E.g. I can prove $(\mathbb{Q}[y])[x]$ is irreducible if I can prove $(\mathbb{Z}[y])[x]$ is irreducible. Assuming this is correct, I proceed as follows.
Let $f(x)=x^3+y^2x^2+3yx-y$ and assume $f(x)$ is reducible in $(\mathbb{Z}[y])[x]$, then we must have $$f(x)=(x^2+ax+b)(x+c),$$ where $a,b,c\in\mathbb{Z}[y]$. This implies that $bc=-y$ which means that $$(b,c)\in\{(1,-y),(-1,y),(-y,1),(y,-1)\}$$ since $b,c\in\mathbb{Z}$, i.e. $x=\pm 1$ or $x=\pm y$.
I took a look at $f(1)$ hoping for a contradiction, but found that everything works out if $y=-1$.
But then I see that when $x=1$ and $y-1$ that the original $f(x)=0$, which means $f(x)$ is reducible, doesn't it?
What's the correct way to proceed? Maybe the wrong approach. Maybe there's a way of applying Eisenstein's criterion...