Step 1: Determine the domain (A) of the function.
For the function $f(x)$ to be defined, the expression inside the square root in the denominator must be strictly positive.
So, $|x| - x^2>0$.
We consider two cases for $|x|$.
Case 1: If $x \ge 0$, the inequality becomes
\[
x - x^2>0 \implies x(1 - x)>0
\]
This is true for $0<x<1$.
Case 2: If $x<0$, the inequality becomes
\[
-x - x^2>0 \implies -x(1 + x)>0 \implies x(x + 1)<0
\]
This is true for $-1<x<0$.
Combining both cases, the domain $A$ is
\[
A = (-1, 0) \cup (0, 1)
\]
Step 2: Determine the range (B) of the function.
Let
\[
y = f(x) = \frac{1}{\sqrt{|x| - x^2}}
\]
Since the denominator involves a square root, $y$ must be positive.
The value of $y$ is minimized when the denominator $\sqrt{|x| - x^2}$ is maximized. Let
\[
g(x) = |x| - x^2
\]
For $x \in (0, 1)$, $g(x) = x - x^2$. This quadratic has a maximum at $x = \frac{1}{2}$, where
\[
g\left(\frac{1}{2}\right) = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}
\]
For $x \in (-1, 0)$, $g(x) = -x - x^2$. This quadratic has a maximum at $x = -\frac{1}{2}$, where
\[
g\left(-\frac{1}{2}\right) = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}
\]
The maximum value of the expression in the denominator is
\[
\sqrt{\frac{1}{4}} = \frac{1}{2}
\]
Hence, the minimum value of $y$ is
\[
y_{\min} = \frac{1}{1/2} = 2
\]
As $x$ approaches the boundaries of the domain ($-1$, $0$, $1$), the expression $|x| - x^2 \to 0^+$, so $y \to \infty$.
Therefore, the range $B$ is
\[
B = [2, \infty)
\]
Step 3: Calculate the union $A \cup B$.
\[
A = (-1, 0) \cup (0, 1)
\]
\[
B = [2, \infty)
\]
\[
\boxed{A \cup B = (-1, 0) \cup (0, 1) \cup [2, \infty)}
\]