6. If $\sqrt{x}-\frac{1}{\sqrt{x}}=-2$, then $x^{2}-\frac{1}{x^{2}}=$
6. $-24 \sqrt{2}$.
From the given, we know $0<x<1$, and
$$
\begin{array}{l}
\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^{2}=x-2+\frac{1}{x}=4 \\
\Rightarrow x+\frac{1}{x}=6 . \\
\text { Also, }\left(x-\frac{1}{x}\right)^{2}=\left(x+\frac{1}{x}\right)^{2}-4=32, \text { so } \\
x-\frac{1}{x}= \pm 4 \sqrt{2} \text { (positive value discarded). }
\end{array}
$$

Then $x^{2}-\frac{1}{x^{2}}=\left(x+\frac{1}{x}\right)\left(x-\frac{1}{x}\right)=-24 \sqrt{2}$.