Nhận xét: $x,y>0$.
Hệ đã cho tương đương với:
$\left\{\begin{array}{l}1-\dfrac{12}{y+3x}=\dfrac{2}{\sqrt x}\\1+\dfrac{12}{y+3x}=\dfrac{6}{\sqrt y}\end{array}\right.$
$\Leftrightarrow \left\{\begin{array}{l}1=\dfrac{1}{\sqrt x}+\dfrac{3}{\sqrt y}\\\dfrac{12}{y+3x}=\dfrac{3}{\sqrt y}-\dfrac{1}{\sqrt x}\end{array}\right.$
$\Leftrightarrow \left\{\begin{array}{l}1=\dfrac{1}{\sqrt x}+\dfrac{3}{\sqrt y}\\\dfrac{12}{y+3x}=(\dfrac{3}{\sqrt y}-\dfrac{1}{\sqrt x})(\dfrac{1}{\sqrt x}+\dfrac{3}{\sqrt y})\end{array}\right.$
$\Leftrightarrow \left\{\begin{array}{l}1=\dfrac{1}{\sqrt x}+\dfrac{3}{\sqrt y}\\\dfrac{12}{y+3x}=\dfrac{9}{y}-\dfrac{1}{x}\end{array}\right.$
$\Leftrightarrow \left\{\begin{array}{l}1=\dfrac{1}{\sqrt x}+\dfrac{3}{\sqrt y}\\27x^2-6xy-y^2=0\end{array}\right.$
$\Leftrightarrow \left\{\begin{array}{l}1=\dfrac{1}{\sqrt x}+\dfrac{3}{\sqrt y}\\(9x+y)(3x-y)=0\end{array}\right.$
$\Leftrightarrow \left\{\begin{array}{l}1=\dfrac{1}{\sqrt x}+\dfrac{3}{\sqrt y}\\y=3x\end{array}\right.$ (vì $9x+y>0$)
$\Leftrightarrow \left\{\begin{array}{l}x=2(2+\sqrt3)\\y=6(2+\sqrt3)\end{array}\right.$