Is $ I=(x+y,x-y)$ a prime ideal in $\mathbb{Z}[x,y]$
Since $x+y, x-y \in \mathbb{Z}$, therefore $(2x,2y) \in (x+y, x-y)$, and this implies that in $R=\frac{\mathbb{Z}[x,y]}{(x-y,x+y)}$, we have $2x=0$. But neither $2$ nor $x$ is zero in $R$. This shows that R is not an integral domain, which implies that $I$ is not a prime ideal.
The idea of the proof is good, and everything seems correct. However, I would personally prefer some more justification of the statement that neither $2$ nor $x$ are zero in $R$. Or, equivalently (and a bit easier in my opinion), why they aren't in $(x+y,x-y)$ as elements of $\Bbb Z[x,y]$. Because in general, this is not easy to just see, and it's not always that easy to spot whether it's easy to see. So being thorough just for safety's sake is usually a good thing.
For instance, you have evaluation homomorphisms $\Bbb Z[x,y]\to \Bbb Z$. The evaluation at $(0,0)$ takes any element in the ideal to $0$ (no element of the ideal has a constant term), but doesn't take $2$ to $0$, so $2$ isn't in the ideal. And the evaluation homomorphism at $(1,1)$ takes any element in the ideal to an even number (no element in the ideal has an odd sum of coefficients), but takes $x$ to $1$, so $x$ can't be in the ideal either.