Given,

$9^{1-{\tan }^{2}x}+9^{{\tan }^{2}x}=10$

Now let $9^{{\tan }^{2}x}=t$, then above equation will be,

$\frac{9}{t}+t=10$

$\Rightarrow t^2-10t+9=0$

$\Rightarrow t=9 \text{or} t=1$

So, when $9^{{\tan }^{2}x}=9\Rightarrow {\tan }^{2}x=1$

$\Rightarrow \tan x=\pm 1\Rightarrow x=\pm \frac{\mathrm{\pi }}{4}$, as given $x\in \left(\frac{-\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right)$

Now when $9^{{\tan }^{2}x}=1$

$\Rightarrow {\tan }^{2}x=0\Rightarrow x=0$

Hence, $\beta =\sum _{x\in S}\left(\frac{x}{3}\right)={\tan }^2\left(\frac{0}{3}\right)+{\tan }^2\left(\frac{\mathrm{\pi }}{12}\right)+{\tan }^2\left(\frac{-\mathrm{\pi }}{12}\right)$

$=0+2{\left(2-\sqrt{3}\right)}^2$

$=14-8\sqrt{3}$

So, the value of $\frac{1}{6}{\left(\beta -14\right)}^2=\frac{64\times 3}{6}=32$

Hence this is the correct option.