Answer: \((3,0)\) and \(\left(-\frac95,-\frac{16}{5}\right)\).

Substitute

\(y=\frac23x-2\)

into

\(x^2+xy=9.\)

This gives

\(x^2+x\left(\frac23x-2\right)=9.\)

Multiplying by \(3\),

\(3x^2+2x^2-6x=27.\)

So

\(5x^2-6x-27=0.\)

Factorising,

\((x-3)(5x+9)=0.\)

Hence

\(x=3\quad\text{or}\quad x=-\frac95.\)

When \(x=3\),

\(y=\frac23(3)-2=0.\)

When \(x=-\frac95\),

\(y=\frac23\left(-\frac95\right)-2=-\frac65-2=-\frac{16}{5}.\)

The points of intersection are

\((3,0)\quad\text{and}\quad \left(-\frac95,-\frac{16}{5}\right).\)