\(\left\{ {\begin{array}{*{20}{c}}{{x^2}y + 2y-{y^2}-2xy = 0,}\\{{{\log }_x}y + 2{{\log }_y}x = 3.\,\,\,\,\,\,\,\,\,}\end{array}} \right.\)
Запишем ОДЗ: \(\left\{ {\begin{array}{*{20}{c}}{x\, \in \,\left( {0;1} \right) \cup \left( {1; + \infty } \right),}\\{y\, \in \,\left( {0;1} \right) \cup \left( {1; + \infty } \right).}\end{array}} \right.\)
Рассмотрим второе уравнение. Воспользуемся свойством: \({\log _a}b = \dfrac{1}{{{{\log }_b}a}}:\)
\({\log _x}y + 2{\log _y}x = 3\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,{\log _x}y + \dfrac{2}{{{{\log }_x}y}} = 3\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left[ \begin{array}{l}{\log _x}y = 1,\\{\log _x}y = 2\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left[ \begin{array}{l}y = x,\\y = {x^2}.\end{array} \right.\)
Тогда исходная система уравнений с учётом ОДЗ равносильна совокупности двух систем:
\(\left[ \begin{array}{l}\left\{ {\begin{array}{*{20}{c}}{y = x,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{{x^2}y + 2y-{y^2}-2xy = 0,}\end{array}} \right.\\\left\{ {\begin{array}{*{20}{c}}{y = {x^2},\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{{x^2}y + 2y-{y^2}-2xy = 0}\end{array}} \right.\end{array} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left[ \begin{array}{l}\left\{ {\begin{array}{*{20}{c}}{y = x,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{{x^3}-3{x^2} + 2x = 0,}\end{array}} \right.\\\left\{ {\begin{array}{*{20}{c}}{y = {x^2},\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{2{x^2}-2{x^3} = 0}\end{array}} \right.\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left[ \begin{array}{l}\left\{ \begin{array}{l}x = 0,\\y = 0,\end{array} \right.\\\left\{ \begin{array}{l}x = 1,\\y = 1,\end{array} \right.\\\left\{ \begin{array}{l}x = 2,\\y = 2.\end{array} \right.\end{array} \right.\)
Из найденных решений только \(\left( {2;2} \right)\) удовлетворяет ОДЗ.
Ответ: \(\left( {2;2} \right).\)