Why not changing bases?

$\frac{4}{\log_3(x)^2-2}+ \log_x \frac{1}{9}>1$ $\Rightarrow \frac{2}{\log_3 x-1}-\frac{2}{\log_3x}>1$

Now putting $\log_3 x=y$ we get $\frac{2}{y-1}-\frac{2}{y}-1>0$ $\Rightarrow \frac{(y-2)(y+1)}{y(y-1)}<0$ $\Rightarrow y \in (-1,0)\cup (1,2)$ $\Rightarrow x \in (\frac{1}{3},1) \cup (3,9)$ Answer is $1+1+3+9=\boxed{14}$