Q:

Why is $y^2 - x$ irreducible in $\mathbb{Q}[x,y]$?

Why is $y^2 - x$ irreducible in $\mathbb{Q}[x,y]$?
Really not sure what technique to use here since the polynomial is multivariate. 

A:

In my opinion, the simplest approach is

If an element is prime, it is irreducible
An element is prime if and only if the ideal it generates is prime
An ideal is prime if and only if the corresponding quotient ring is a domain

and compute the quotient ring
$$ \mathbb{Q}[x,y] / (y^2 - x) $$
(It may help to note that $y^2 - x$ is a monic, linear, univariate polynomial over the coefficient ring $\mathbb{Q}[y]$. This note can help with other approaches too)

A:

DirectSuperHint:
If $y^2 - x$ is reducible, then there exists polynomials $p,q \in \mathbb{Q}[x,y]$ s.t $y^2 - x = pq$, now the degree of this polynomials must be one, i.e, $p = a+bx+cy,q=d + ex + fy$ with $a,b,c,d,e,f \in\mathbb{Q}$. Do the product and equals the coeficients with $y^2-x$ (a system of equations), now you will obtain a contradiction. (Possibly $a=b=c=d=e=f=0$).

