jmc

$x$ is not in the domain of $f$ if the denominator is zero. Since both absolute values are nonnegative, both must be zero for the denominator to be zero. So

$$\begin{array}{rl}0& =x^2-6x+8=(x-2)(x-4)\Rightarrow x=2\text{ or }x=4.\\ 0& =x^2+x-6=(x-2)(x+3)\Rightarrow x=2\text{ or }x=-3.\end{array}$$The only value of $x$ which makes both absolute values zero is $x=\boxed{2}$.