Let $y=\frac{9 \cdot 3^{2 x}+6 \cdot 3^x+4}{9 \cdot 3^{2 x}-6 \cdot 3^x+4}$
Put, $3^x=t$
$\therefore \quad y=\frac{9 \cdot t^2+6 t+4}{9 t^2-6 t+4}$
$\Rightarrow y\left(9 t^2-6 t+4\right)=9 t^2+6 t+4$
$\Rightarrow 9 t^2(y-1)-6 t(y+1)+(4 y-4)=0$
$\Rightarrow \quad(y-1) t^2 \frac{-6}{9}(y+1) t+\frac{4}{9}(y-1)=0$
$\Rightarrow(y-1) t^2-\frac{2}{3}(y+1) t+\frac{4}{9}(y-1)=0$
$\because t$ is real
$\therefore D \geq 0$
$\Rightarrow \quad \frac{4}{9}(y+1)^2-4(y-1) \frac{4}{9}(y-1) \geq 0$
$\Rightarrow \quad \frac{4}{9}(y+1)^2-\frac{16}{9}(y-1)^2 \geq 0$
$\Rightarrow \quad(y+1)^2-4(y-1)^2 \geq 0$
$\Rightarrow \quad y^2+1+2 y-4 y^2-4+8 y \geq 0$
$\Rightarrow \quad-3 y^2+10 y-3 \geq 0$
$\Rightarrow \quad 3 y^2-10 y+3 \leq 0$
$\Rightarrow \quad(y-3)(3 y-1) \leq 0 \Rightarrow \frac{1}{3} \leq y \leq 3$
Hence, minimum value $=\frac{1}{3}$.