jmc

We can write $$f(x)=\sqrt{x(8-x)}-\sqrt{(x-6)(8-x)}.$$Thus, $x$ is defined only for $6\le x\le 8.$ Then $$\begin{array}{rl}f(x)& =\sqrt{8-x}(\sqrt{x}-\sqrt{x-6})\\ & =\sqrt{8-x}\cdot \frac{(\sqrt{x}-\sqrt{x-6})(\sqrt{x}+\sqrt{x-6})}{\sqrt{x}+\sqrt{x-6}}\\ & =\sqrt{8-x}\cdot \frac{x-(x-6)}{\sqrt{x}+\sqrt{x-6}}\\ & =\sqrt{8-x}\cdot \frac{6}{\sqrt{x}+\sqrt{x-6}}.\end{array}$$On the interval $6\le x\le 8,$ $\sqrt{8-x}$ is decreasing, and $\sqrt{x}+\sqrt{x-6}$ is increasing, which means $\frac{6}{\sqrt{x}+\sqrt{x-6}}$ is decreasing. Therefore, the maximum value of $f(x)$ is $$f(6)=\sqrt{2}\cdot \frac{6}{\sqrt{6}}=\boxed{2\sqrt{3}}.$$