smc

We can rewrite the function as $f(x)=\sqrt{x(8-x)}-\sqrt{(x-6)(8-x)}$ and then factor it to get $f(x)=\sqrt{8-x}\left(\sqrt{x}-\sqrt{x-6}\right)$. From the expressions under the square roots, it is clear that $f(x)$ is only defined on the interval $[6,8]$. The $\sqrt{8-x}$ factor is decreasing on the interval. The behavior of the $\sqrt{x}-\sqrt{x-6}$ factor is not immediately clear. But rationalizing the numerator, we find that $\sqrt{x}-\sqrt{x-6}=\frac{6}{\sqrt{x}+\sqrt{x-6}}$, which is monotonically decreasing. Since both factors are always positive, $f(x)$ is also positive. Therefore, $f(x)$ is decreasing on $[6,8]$, and the maximum value occurs at $x=6$. Plugging in, we find that the maximum value is $\boxed{2\sqrt{3}}$.