irreducibility of a polynomial in several variables over ANY field The irreeducibility of a polynomial $f\!\in\!K[x_1,\ldots,x_n]$ in general depends on what the field $K$ is (for example, if $K=\mathbb{R}$, then $f=x_1^2+1$ is irreducible, but if $K=\mathbb{C}$, it is not). 
However, when $n\geq2$, there exist polynomials that are irreducible no matter what $K$ is (for example $x_i-x_j^d$ where $d\in\mathbb{N}$ is arbitrary and $i\neq j$, is always irreducible).
Question: Is there a command in any computer algebra system (preferably SINGULAR), that could (in some cases) confirm if a polynomial is irreducible over any field?
For example, in A Singular Introduction to Commutative Algebra, page 222, there is written:

Since there is mention of the quotient field $Q(A)$, this means $A$ is a domain, i.e. $x^4+6x^2y-y^3$ is irreducible.
Question: How can I show that $x^4+6x^2y-y^3$ is irreducible over any $K$?
 A: This is a minor variation of Georges's answer. Assume by contradiction that $y^3-6x^2y-x^4$ is reducible in $K[x,y]$. Then we necessarily have 
$$
y^3-6x^2y-x^4=\Big(y-f(x)\Big)\Big(y^2+g(x)\, y+h(x)\Big)
$$
for some $f(x),g(x),h(x)$ in $K[x]$. This implies in particular $x^4=f(x)\,h(x)$. Thus $f(x)=a\, x^n$ with $0\le n\le4$ and $a\in K^\times$, and we easily rule out the five "possible" values of $n$.
(I upvoted the three other posts.)
EDIT. In order to prove that $y^3-6x^2y-x^4$ is irreducible in $K[x,y]$, it suffices to assume that $K$ is a domain. Indeed, in the above argument, we only need to know that (up to association) 
$$
x^4=x\cdot x\cdot x\cdot x
$$
is the only decomposition of $x^4$ as a product of irreducibles in $K[x]$.
A: Let me show in a human(e) way that $f(y)=y^3-6x^2y-x^4=-\text {(your  polynomial)} \quad $ is irreducible over $K$.  
Let $A=K[x]$ and $L=Frac(A)=K(x)$.
 We have to prove that $f(y)\in K[x,y]=A[y]$  is irreducible or equivalently [since $f(y)$ is monic] that $f(y)$ is irreducible over $L$ or [since $f(y)$ has degree $3$] that $f(y)$ has no root in $L=K(x)$.
However , by normality of $A=K[x]$, such a root would have to be in $A$ and moreover divide $-x^4$. Since obviously no such root exists, $f(y)$ is irreducible as a polynomial in  $A[y]$, in $L[y]$ or (as originally required) in $K[x,y]$.
A: I'll answer your 2nd question, as I think the 1st one is off-topic.
Since you know that there is an isomorphism as the one underlined in red in your image, you know that the quotient $A$ of $K[x,y]$ by the ideal generated by your polinomial is isomorphic to a subring of $K[t]$.
Every subring of $K[t]$ is a domain, the ring $A$ is itself a domain and your polynomial irreducible,
