What is the domain of $f\left( x,y\right)=\arctan\left(\frac{1}{\sin\left( x+y\right)}\right)$ Originally, I believed this was as simple as the domain being  
$$D\left(f\right)=\left\{\left(x,y\right) \in\mathbb{R}^{2}:\sin\left(x+y\right) \neq 0\right\}  $$  
However, after looking further I found the domain to be
$D\left(f\right)=\left\{\left(x,y\right):2n\pi + x < y < 2\left(n+1\right)\pi + x, n \in\mathbb{Z}\right\}  $
Is this along the right lines?

$\arctan$'s domain is all of $\mathbb{R}$. So you only need worry about when $\frac{1}{\sin(x+y)}$ is undefined. So the domain is $$\{(x,y): \sin(x+y)\neq0\}$$
More explicitly, this is $$\left\{(x,y): \frac{x+y}{\pi}\not\in\mathbb{Z}\right\}$$
The way I have written it, $\frac{x+y}{\pi}$ must fall in between two consecutive integers. 
Your domain, looking with $n=0$, allows $x=-1$, $y=1$. Which is clearly not in the domain. 
