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There are two things you want to avoid in the domain of $f$.
1) $x^2+y^2+4<0$
2) $0=v = y-x$
But $x^2+y^2 +4 \geq 4 >0$ so this shouldn't be a problem. However if $x=y$ then $v=0$ so the domain is $D=\{(x,y)\mid x \neq y\}$.
Edit Here is a "picture" of $D$ as asked in the comment. $D$ is basically everything except the points that are on the red line. Note that this picture only represents a little part of $D$ since it is an unbounded domain (e.g. $(10^{2137612},\pi)$ belongs to $D$ but is not on the picture).
Since $v \neq 0$ by definition, then $y-x \neq 0$ ---> $y \neq x$. Therefore, the line y= x is excluded from the domain of f. Also, since $\sqrt{x^2+y^2+4}\geq 0$ and $y \neq x$, $\sqrt{x^2+y^2+4}$. So the domain is $ D = \{(x,y)| \sqrt{x^2+y^2+4}\geq 0$ where $y\neq x$} Surb gives an excellent sketch of the domain above in his/her reply.