Из третьего уравнения:
\(yz = 3\left( {y + z} \right)\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,yz-3z = 3y\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,z\left( {y-3} \right) = 3y\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,z = \dfrac{{3y}}{{y-3}}.\)
Подставим во второе уравнение:
\(\dfrac{{3xy}}{{y-3}} = 2\left( {x + \dfrac{{3y}}{{y-3}}} \right)\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,3xy = 2\left( {xy-3x + 3y} \right)\,\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\,\,xy = 6y-6x\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,xy = 6\left( {y-x} \right).\)
Вместе с первым уравнением исходной системы получим:
\(\left\{ {\begin{array}{*{20}{c}}{xy = x + y,\,\,\,\,\,\,\,}\\{xy = 6\left( {y-x} \right)}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{xy = x + y,\,\,\,\,\,\,\,\,}\\{\dfrac{{xy}}{{xy}} = \dfrac{{x + y}}{{6\left( {y-x} \right)}}}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{xy = x + y,}\\{y = \dfrac{{7x}}{5}\,\,\,\,\,\,\,\,\,}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{\dfrac{{7{x^2}}}{5} = x + \dfrac{{7x}}{5},}\\{y = \dfrac{{7x}}{5}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{7{x^2}-12x = 0,}\\{y = \dfrac{{7x}}{5}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\left\{ {\begin{array}{*{20}{c}}{\left[ {\begin{array}{*{20}{c}}{x = 0,\,\,\,\,}\\{x = \dfrac{{12}}{7},}\end{array}} \right.}\\{y = \dfrac{{7x}}{5}}\end{array}\,} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{\left\{ {\begin{array}{*{20}{c}}{x = 0,}\\{y = 0,}\end{array}} \right.}\\{\left\{ {\begin{array}{*{20}{c}}{x = \dfrac{{12}}{7},}\\{y = \dfrac{{12}}{5}.}\end{array}} \right.}\end{array}} \right.\)
Если \(x = y = 0\), то \(z = 0;\)
если \(x = \dfrac{{12}}{7},\,\,\,y = \dfrac{{12}}{5}\), то \(\dfrac{{12}}{7}z = 2\left( {\dfrac{{12}}{7} + z} \right)\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,z = -12.\)
Ответ: \(\left( {0;0;0} \right),\,\,\,\left( {\dfrac{{12}}{7};\dfrac{{12}}{5};-12} \right).\)