Answer
$$\left\{ {\left( {\frac{6}{{\sqrt 7 }}, - \frac{{3\sqrt 7 }}{7}} \right),\left( { - \frac{6}{{\sqrt 7 }},\frac{{3\sqrt 7 }}{7}} \right),\left( {\sqrt 3 ,2\sqrt 3 } \right),\left( { - \sqrt 3 {\text{ }}, - 2\sqrt 3 } \right)} \right\}$$
Work Step by Step
$$\eqalign{
& \,\,3{x^2} + 2xy - {y^2} = 9\,\,\,\,\,\,\left( {\bf{1}} \right) \cr
& \,\,\,\,\,{x^2} - \,\,\,xy + \,\,{y^2} = 9\,\,\,\,\left( {\bf{2}} \right) \cr
& \cr
& {\text{Add both equations to obtain the equation }}\left( {\bf{3}} \right) \cr
& \,3{x^2} + 2xy - {y^2} = 9 \cr
& \underline {\,\,\,\,{x^2} - \,\,\,xy + \,\,{y^2} = 9} \cr
& 4{x^2}\, + xy\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 18\,\,\,\,\,\,\,\left( {\bf{3}} \right) \cr
& \cr
& {\text{Solve the equation }}\left( {\bf{3}} \right){\text{ for }}y \cr
& 4{x^2}\, + xy = 18 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,y = \frac{{18 - 4{x^2}}}{x} \cr
& \cr
& {\text{Substitute }}\frac{{18 - 4{x^2}}}{x}{\text{ for }}y{\text{ into the equation }}\left( {\bf{2}} \right) \cr
& {x^2} - \,\,\,x\left( {\frac{{18 - 4{x^2}}}{x}} \right) + \,\,{\left( {\frac{{18 - 4{x^2}}}{x}} \right)^2} = 9 \cr
& {\text{Solve for }}x \cr
& {x^2} - \,\,\,\left( {18 - 4{x^2}} \right) + \,\,\frac{{324 - 144{x^2} + 16{x^4}}}{{{x^2}}} = 9 \cr
& {x^4} - \,\,\,{x^2}\left( {18 - 4{x^2}} \right) + \,\,324 - 144{x^2} + 16{x^4} = 9{x^2} \cr
& {x^4} - 18\,{x^2} + 4{x^4} + \,\,324 - 144{x^2} + 16{x^4} = 9{x^2} \cr
& 21{x^4} - 171{x^2} + 324 = 0 \cr
& \left( {7{x^2} - 36} \right)\left( {{x^2} - 3} \right) = 0 \cr
& 7{x^2} - 36 = 0,\,\,\,\,\,\,\,{x^2} - 3 = 0 \cr
& 7{x^2} = 36,\,\,\,\,\,\,\,{x^2} = 3 \cr
& {x_1} = \frac{6}{{\sqrt 7 }},\,\,\,\,{x_2} = - \frac{6}{{\sqrt 7 }},\,\,\,\,\,\,{x_3} = \sqrt 3 ,\,\,\,\,\,{x_4} = - \sqrt 3 \cr
& \cr
& {\text{Substitute }}{x_1}{\text{ = }}\frac{6}{{\sqrt 7 }}{\text{ for }}x{\text{ into the equation }}y = \frac{{18 - 4{x^2}}}{x} \cr
& y = \frac{{18 - 4{{\left( {\frac{6}{{\sqrt 7 }}} \right)}^2}}}{{\frac{6}{{\sqrt 7 }}}} = - \frac{{3\sqrt 7 }}{7} \cr
& {\text{The first solution is }}\left( {\frac{6}{{\sqrt 7 }}, - \frac{{3\sqrt 7 }}{7}} \right) \cr
& \cr
& {\text{Substitute }}{x_2}{\text{ = }} - \frac{6}{{\sqrt 7 }}{\text{ for }}x{\text{ into the equation }}y = \frac{{18 - 4{x^2}}}{x} \cr
& y = \frac{{18 - 4{{\left( { - \frac{6}{{\sqrt 7 }}} \right)}^2}}}{{ - \frac{6}{{\sqrt 7 }}}} = \frac{{3\sqrt 7 }}{7} \cr
& {\text{The second solution is }}\left( { - \frac{6}{{\sqrt 7 }},\frac{{3\sqrt 7 }}{7}} \right) \cr
& \cr
& {\text{Substitute }}{x_3}{\text{ = }}\sqrt 3 {\text{ for }}x{\text{ into the equation }}y = \frac{{18 - 4{x^2}}}{x} \cr
& y = \frac{{18 - 4{{\left( {\sqrt 3 {\text{ }}} \right)}^2}}}{{\sqrt 3 {\text{ }}}} = 2\sqrt 3 \cr
& {\text{The third solution is }}\left( {\sqrt 3 ,2\sqrt 3 } \right) \cr
& \cr
& {\text{Substitute }}{x_3}{\text{ = }} - \sqrt 3 {\text{ for }}x{\text{ into the equation }}y = \frac{{18 - 4{x^2}}}{x} \cr
& y = \frac{{18 - 4{{\left( { - \sqrt 3 {\text{ }}} \right)}^2}}}{{ - \sqrt 3 {\text{ }}}} = - 2\sqrt 3 \cr
& {\text{The fourth solution is }}\left( { - \sqrt 3 {\text{ }}, - 2\sqrt 3 } \right) \cr
& \cr
& {\text{Therefore, the solution set of the system is}} \cr
& \left\{ {\left( {\frac{6}{{\sqrt 7 }}, - \frac{{3\sqrt 7 }}{7}} \right),\left( { - \frac{6}{{\sqrt 7 }},\frac{{3\sqrt 7 }}{7}} \right),\left( {\sqrt 3 ,2\sqrt 3 } \right),\left( { - \sqrt 3 {\text{ }}, - 2\sqrt 3 } \right)} \right\} \cr} $$
