My approach:

\(\left\{{}\begin{matrix}3x^2+6xy-3y^2=6\\2x^2+2xy+2y^2=6\end{matrix}\right.\)

It implies that

\(\left(3x^2+6xy-3y^2\right)-\left(2x^2+2xy+2y^2\right)=6-6=0\)

\(x^2+4xy-5y^2=0\)

\(\left[{}\begin{matrix}x=y\\x=5y\end{matrix}\right.\)

Case 1: x=y, subtitute to (1) we have

\(x^2=1\) \(\Leftrightarrow\) \(x=\pm1\)

Then we get