Answer
(a) All points in 2-space on or between the vertical lines $x = ±2$
(b) All points in 2-space above the line $y = 2x$
(c) All points in 3-space not on the plane $x + y + z = 0$
Work Step by Step
(a) $f(x, y) = \frac{\sqrt (4-x^2)}{y^2+3}$
First, we find the domain;
$4-x^2 \geq 0$ [NB: values under the square root must not be negative]
$x^2\geq 4 => x \geq _{-}^+2$
$y^2 +3 \ne 0$ [NB: denomenator must not be zero]
$y^2\ne -3$
Domain = {$x, y \in R$ and $x \geq _{-}^+2$ and $y^2\ne -3$}
(b) $f(x, y) = \ln(y-2x)$
First, we find the domain;
$y-2x\gt0$ [NB: values under log must not be zero or negative]
$y\gt2x$
Domain = {$x, y \in R$ and $y\gt2x$}
(c) $f(x, y, z) = \frac{xyz}{x+y+z}$
First, we find the domain;
$x+y+z \ne 0$ [NB: denomenator must not be zero]
Domain = {$x, y, z \in R$ and $x+y+z \ne 0$}
Therefore the domains help us to describe the domain in words
