3. $2 \sqrt{3}$.
$\because y=\sqrt{(8-x) x}-\sqrt{(8-x)(x-6)}$ is a real number if and only if $6 \leqslant x \leqslant 8$, and
$$\begin{aligned} y= & \sqrt{8-x}(\sqrt{x}-\sqrt{x-6})=\sqrt{8-x} \\ & \cdot \frac{(\sqrt{x}-\sqrt{x-6})(\sqrt{x}+\sqrt{x-6})}{\sqrt{x}+\sqrt{x-6}} \\ = & \frac{6 \sqrt{8-x}}{\sqrt{x}+\sqrt{x-6}} \end{aligned}$$

holds for all $6 \leqslant x \leqslant 8$.
When $x=6$, the numerator of the last expression is maximized, and the denominator is minimized. Therefore, the value of the fraction is maximized, which is $\frac{6 \sqrt{2}}{\sqrt{6}}=2 \sqrt{3}$.