$\begin{aligned}
& \text { Let } y=\frac{x^2+14 x+9}{x^2+2 x+3} \\
& \Rightarrow \quad x^2 y+2 x y+3 y=x^2+14 x+9 \\
& \Rightarrow \quad x^2(y-1)+2 x(y-7)+3 y-9=0
\end{aligned}$
Here, $x \in \mathbf{R}$
$\begin{aligned}
\therefore \quad 4(y-7)^2-4(y-1)(3 y-9) & \geq 0 \\
(y-7)^2-\left(3 y^2-12 y+9\right) & \geq 0 \\
y^2-14 y+49-3 y^2+12 y-9 & \geq 0 \\
\Rightarrow \quad 2 y^2+2 y-40 & \leq 0 \Rightarrow y^2+y-20 \leq 0 \\
(y+5)(y-4) & \leq 0 \quad \therefore \quad y \in[-5,4]
\end{aligned}$
$\therefore$ Maximum and minimum value of $\frac{x^2+14 x+9}{x^2+2 x+3}$ are 4 and -5 respectively.