Điều kiện: $\color{red}{x\neq -3.}$
$\color{green}{x^2+\frac{9x^2}{(x+3)^2}=7}$$\color{green}{\Leftrightarrow (x-\frac{3x}{x+3})^2+6.\frac{x^2}{x+3}=7}$
$\color{green}{\Leftrightarrow (\frac{x^2}{x+3})^2+6.\frac{x^2}{x+3}-7=0}$ $\color{green}{(\bigstar)}$
Đặt $\color{red}{y=\frac{x^2}{x+3}}$, phương trình $\color{green}{(\bigstar)}$ trở thành:
$\color{red}{y^2+6y-7=0}$
$\color{red}{\Leftrightarrow \left[\ \begin{array}{l} y=1\\ y=-7 \end{array} \right.\Leftrightarrow \left[\ \begin{array}{l} x^2=x+3\\ x^2+7x+21=0(vn) \end{array} \right.\Leftrightarrow x=\frac{1 \pm \sqrt {13}}{2}(tm)}$