2. Find all pairs of positive numbers $(x, y)$ that satisfy the system of equations $\left\{\begin{array}{l}3 y-\sqrt{\frac{y}{x}}-6 \sqrt{x y}+2=0, \\ x^{2}+81 x^{2} y^{4}=2 y^{2} .\end{array}\right.$ Answer: $\left(\frac{1}{3} ; \frac{1}{3}\right),\left(\frac{\sqrt[4]{31}}{12} ; \frac{\sqrt[4]{31}}{3}\right)$
Solution. Let $\sqrt{\frac{y}{x}}=u, \sqrt{x y}=v \quad$ (with $u>0, \quad v>0$). Then $\quad u v=\sqrt{\frac{y}{x}} \cdot \sqrt{x y}=\sqrt{y^{2}}=|y|=y$, $\frac{v}{u}=\sqrt{x y}: \sqrt{\frac{y}{x}}=\sqrt{x^{2}}=|x|=x$, since by the condition $x$ and $y$ are positive. The system takes the form

$$
\left\{\begin{array} { l } 
{ 3 u v - u - 6 v + 2 = 0 } \\
{ \frac { v ^ { 2 } } { u ^ { 2 } } + 8 1 v ^ { 6 } u ^ { 2 } = 2 u ^ { 2 } v ^ { 2 } }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
(u-2)(3 v-1)=0 \\
1+81 v^{4} u^{4}=2 u^{4}
\end{array}\right.\right.
$$

From the first equation, it follows that $u=2$ or $v=\frac{1}{3}$.

If $u=2$, then $1+81 \cdot 16 v^{4}=32$, from which $v=\frac{\sqrt[4]{31}}{6}$; then $x=\frac{v}{u}=\frac{\sqrt[4]{31}}{12}, y=u v=\frac{\sqrt[4]{31}}{3}$.

If $v=\frac{1}{3}$, then $1+u^{4}=2 u^{4}$, from which $u=1$; then $x=\frac{v}{u}=\frac{1}{3}, y=u v=\frac{1}{3}$.