Given that, for all real $x$ , the expression $\frac{x^2-2x+4}{x^2+2x+4}$ lies between $\frac{1}{3}$ and 3. The values between which the expression $\frac{9\cdot 3^{2x}+6\cdot 3^x+4}{9\cdot 3^{2x}-6\cdot 3^x+4}$ lies are

Given  that $\frac{1}{3}<\frac{x^2-2x+4}{x^2+2x+4}<3$  for all $x\in R$ 

$\Rightarrow \frac{1}{3}<\frac{x^2+2x+4}{x^2-2x+4}<3$ for all $x\in R$       ….…(1)

Let $3^{x+1}=y$

Then $y\in R$ for all $x\in R$ .

$\therefore \frac{9\cdot 3^{2x}+6\cdot 3^x+4}{9\cdot 3^{2x}-6\cdot 3^x+4}=\frac{3^2{.3}^{2x}+{\mathrm{2.3.3}}^x+4}{3^2{.3}^{2x}-{\mathrm{2.3.3}}^x+4}$

$=\frac{3^{2x+2}+2\cdot 3^{x+1}+4}{3^{2x+2}-2\cdot 3^{x+1}+4}$

.$=\frac{y^2+2y+4}{y^2-2y+4}$

Form (1), $\frac{1}{3}<\frac{y^2+2y+4}{y^2-2y+4}<3$

$\therefore \frac{1}{3}<\frac{9\cdot 3^{2x}+6\cdot 3^x+4}{9\cdot 3^{2x}-6\cdot 3^x+4}<3$