Answer
Domain = $\{(x,y)|\quad y\geq x^{2},\quad x\neq\pm 1\}$
(see image)
Work Step by Step
Restriction on the denominator:
f is not defined when $1-x^{2}=0,$ that is, when $x=\pm 1$
(the two vertical lines are excluded from the domain)
Restriction on the square root: f IS defined only for
$y-x^{2}\geq 0$
$y\geq x^{2}$
The plane $\mathbb{R}^{2}$ is divided into two regions by the parabola $y=x^{2}.$
The point (0,1) satisfies the inequality, so the solution region contains this point.
The parabola itself belongs to the solution set.
Combining the two restrictions, the domain contains all points between the "wings" of the parabola, except the points lying on the vertical lines $x=\pm 1.$
Domain = $\{(x,y)|\quad y\geq x^{2},\quad x\neq\pm 1\}$
(see image)
