[tex] \frac{ x^{2 }-9 }{9 x^{2} - 1} : \frac{x - 3}{3x + 1}= \\\\9x^2-1\neq 0 \ \ and \ \ 3x+1 \neq 0 \ \ and\ \ x-3\neq 0 \\ \\ (3x-1)(3x+1)\neq 0 \ \ and \ \ 3x \neq -1 \ \ and\ \ x \neq 3\\\\3x-1 \neq 0 \ \ and \ \3x+1 \neq 0 \ \ and \ \ 3x \neq -1 \ \ and\ \ x \neq 3\\\\3x \neq 1 \ \ and \ \3x \neq -1 \ \ and \ \ 3x \neq- \frac{ 1}{3} \ \ and\ \ x \neq 3\\\\ x \neq\frac{ 1}{3} \ \ and \ \ x \neq -\frac{1}{3} \ \ and \ \ 3x \neq- \frac{ 1}{3} \ \ and\ \ x \neq 3 [/tex]
[tex]D=R\setminus \left \{ -\frac{1}{3},\frac{1}{3},3 \right \}\\\\\\\frac{ x^{2 }-9 }{9 x^{2} - 1} : \frac{x - 3}{3x + 1}=\frac{( x-3)(x+3) }{(3 x - 1)(3x+1)} \cdot \frac{3x + 1}{x - 3}=\frac{x+3}{3x-1}[/tex]
