imc

$x♣y=x^{2}y-x{y}^2$ and $y♣x=y^{2}x-y{x}^2$. Therefore, we have the equation $x^{2}y-x{y}^2=y^{2}x-y{x}^2$ Factoring out a $-1$ gives $x^{2}y-x{y}^2=-(x^{2}y-x{y}^2)$ Factoring both sides further, $xy(x-y)=-xy(x-y)$. It follows that if $x=0$, $y=0$, or $(x-y)=0$, both sides of the equation equal 0. By this, there are 3 lines ($x=0$, $y=0$, or $x=y$) so the answer is $\boxed{\text{three lines}}$.