Let $h(x,y,z) = (z^2 -xz + zy -xy)^{1/4}$. What is the domain on this function?
I know that \begin{align*} z^2 -xz + zy -xy \geq 0 \\ \implies z(x+y) -x(z+y) \geq 0 \\ \implies (z-x)(z+y) \geq 0 \end{align*}
So is $D(h) = \{ (x,y,z) \in \mathbb{R}^3 | (z-x)(z+y) \geq 0 \}$ sufficient?
Also how can this be described in words? Is it just a pair of intersecting planes?
Remember that for $a,b \in \mathbb{R}$
$ab \geq 0 \Rightarrow a \geq 0\ \ \text{and}\ \ b\geq 0 \ \ \text{or} \ \ a \leq 0 \ \ \text{and} \ \ b \leq 0$
Here is a plot in Mathematica of it. Using range from $(-2,2)$ for every coordinate.
$\hskip1.5in$
Your $D(h)$ is alright. Whether a point $(x,y,z)$ belongs to $D(h)$ depends on the signs of $z-x$ and $z+y$. Now $z-x=0$ and $z+y=0$ each define a plane in ${\mathbb R}^3$. When a moving point crosses one of these planes the point changes its "belonging status" versus $D(h)$. The two planes intersect in the line $$g:\quad t\mapsto (t,-t,t)\qquad(-\infty<t<\infty)$$ and partition ${\mathbb R}^3$ into four "wedges" meeting in $g$. The domain $D(h)$ is the union of two non-adjacent wedges of these four: We want that $z-x$ and $z+y$ don't have opposite signs. (Unfortunately this is difficult to draw in a nice figure showing the axes, the planes, and $g$.)
