The domain of the function   $\mathrm{f}\left(\mathrm{x}\right)=\frac{{\cos }^{-1}\left(\frac{{\mathrm{x}}^2-5\mathrm{x}+6}{{\mathrm{x}}^2-9}\right)}{{\log }_{\mathrm{e}}\left({\mathrm{x}}^2-3\mathrm{x}+2\right)}$ is 

$\begin{array}{l}-1\le \frac{{\mathrm{x}}^2-5\mathrm{x}+6}{{\mathrm{x}}^2-9}\le 1\\ \frac{{\mathrm{x}}^2-5\mathrm{x}+6}{{\mathrm{x}}^2-9}-1\le 0\end{array}$
$\begin{array}{l}\frac{1}{x+3}\ge 0\\ x\in (-3,\mathrm{\infty })\dots \dots \left(1\right)\\ \frac{x^2-5x+6}{x^2-9}+1\ge 0\\ \frac{2x+1}{x+3}\ge 0\\ x\in (-\mathrm{\infty },-3)\cup \left(-\frac{1}{2},\mathrm{\infty }\right)....\left(2\right)\end{array}$
after taking intersection
$\begin{array}{l}\mathrm{x}\in \left[-\frac{1}{2},\mathrm{\infty }\right)\\ {\mathrm{x}}^2-3\mathrm{x}+2>0\\ \mathrm{x}\in (-\mathrm{\infty },1)\cup (2,\mathrm{\infty })\\ {\mathrm{x}}^2-3\mathrm{x}+2\ne 1\\ \mathrm{x}\ne \frac{3\pm \sqrt{5}}{2}\end{array}$
after taking intersection of each solution
$\left[-\frac{1}{2},1\right)\cup (2,\mathrm{\infty })-\left\{\frac{3+\sqrt{5}}{2},\frac{3-\sqrt{5}}{2}\right\}$