3. Let $x$ be a real number. Then the maximum value of the function $y=\sqrt{8 x-x^{2}}-$ $\sqrt{14 x-x^{2}-48}$ is
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 $\frac{6 \sqrt{2}}{\sqrt{6}}=2 \sqrt{3}$.