Answer
(a) The domain is $(x,y)\in\mathbb{R}^2$. The range is $f(x,y)\in \mathbb{R}$.
(b) The points at which $f(x,y)=0$ are given by $(0,y), y\in \mathbb{R}.$
(c) It does not pass through all octants.
Work Step by Step
The given function is $$f(x,y) =\frac{-4x}{x^2+y^2+1}.$$
(a) For domain we have only one constraint, the denominator must not be zero:
$x^2+y^2+1\neq 0$. However, because $x^2+y^2$ in never negative this constraint is satisfied for any real $x$ and $y$ (whichever $x$ and $y$ we take the expression $x^2+y^2+1$ is never $0$). This means that the domain for both $x$ and $y$ is the set of real numbers so we write $(x,y)\in\mathbb{R^2}$.
For range, we will have all real numbers because in the numerator $-4x$ takes all real values, and then we divide by some positive number.
(b) We set $f(x,y)=0$ which is
$$\frac{-4x}{x^2+y^2+1} =0$$
and that happens only when $x=0$. $y$ can be anything so the points in the $xy$ plane are given by $(0,y)$, $y\in\mathbb{R}.$
(c) This function does not pass through the first octant where $x$, $y$ and $z$ (which is used to represent the function) are all positive. This is because when $x$ and $y$ are positive the value of the function is always negative (the numerator $-4x$ is negative and the denominator is positive).
