5. Find all pairs of positive numbers $(x, y)$ that satisfy the system of equations $\left\{\begin{array}{l}x-3 \sqrt{x y}-2 \sqrt{\frac{x}{y}}+6=0, \\ x^{2} y^{2}+x^{4}=82 .\end{array}\right.$ Answer: $\left(3 ; \frac{1}{3}\right),\left(\sqrt[4]{66} ; \frac{4}{\sqrt[4]{66}}\right)$
Solution. Let $\sqrt{\frac{x}{y}}=u, \quad \sqrt{x y}=v \quad$ (with $\quad u>0, \quad v>0$ ). Then $\quad u v=\sqrt{\frac{x}{y}} \cdot \sqrt{x y}=\sqrt{x^{2}}=|x|=x$, $\frac{v}{u}=\sqrt{x y}: \sqrt{\frac{x}{y}}=\sqrt{y^{2}}=|y|=y$, since $x$ and $y$ are positive by condition. The system takes the form

$$
\left\{\begin{array} { l } 
{ u v - 3 v - 2 u + 6 = 0 , } \\
{ v ^ { 4 } + u ^ { 4 } v ^ { 4 } = 8 2 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
(v-2)(u-3)=0 \\
v^{4}+u^{4} v^{4}=82
\end{array}\right.\right.
$$

From the first equation, it follows that $v=2$ or $u=3$.

If $v=2$, then $16+16 u^{4}=82$, from which $u=\frac{\sqrt[4]{66}}{2}$; then $y=\frac{v}{u}=\frac{4}{\sqrt[4]{66}}, x=u v=\sqrt[4]{66}$.

If $u=3$, then $v^{4}+81 v^{4}=82$, from which $v=1$; then $y=\frac{v}{u}=\frac{1}{3}, x=u v=3$.