The domain of the function $$f(x) = \dfrac{1}{\sqrt{|x| - x}}$$ is:

The given function is $$f(x)=\dfrac{1}{\sqrt{|x|-x}}$$.

For any real-valued function that contains a square root in the denominator, both of the following conditions must hold:

1. The radicand (the quantity inside the square root) must be strictly positive: $$|x|-x \gt 0$$,
2. The denominator itself must be non-zero, which is already ensured if the radicand is positive.

Evaluate the radicand case-wise because of the absolute value:

Then $$|x| = x$$, so $$|x|-x = x-x = 0$$. The inequality $$0 \gt 0$$ is impossible; hence no $$x \ge 0$$ satisfies the condition.

Now $$|x| = -x$$ (since $$x$$ is negative). Therefore,

$$|x|-x = (-x)-x = -2x.$$

Because $$x \lt 0$$, we have $$-2x \gt 0$$ automatically. Thus every negative real number satisfies the required inequality.

Combining the two cases, the only allowable values are all real numbers less than zero:

$$\boxed{(-\infty,\,0)}$$.

Hence the domain matches Option B: $$(-\infty, 0)$$.