If one of the lines of $m{y}^2+(1-m^2)xy-m{x}^2=0$ is a bisector of the angle between the lines $xy=0$, then $m$ is:

The lines $xy=0$ are the $x$ -axis ($y=0$) and $y$ -axis ($x=0$).The angle bisectors of these axes are $y=x$ and $y=-x$.If $y=x$ is a line of the given equation:Substitute $y=x$ into $m{y}^2+(1-m^2)xy-m{x}^2=0$:
$$m{x}^2+(1-m^2)x^2-m{x}^2=0$$
$$(1-m^2)x^2=0⟹m^2=1⟹m=\pm 1$$
If $y=-x$ is a line:
$$m(-x{)}^2+(1-m^2)x(-x)-m{x}^2=0$$
$$m{x}^2-(1-m^2)x^2-m{x}^2=0$$
$$-(1-m^2)x^2=0⟹m=\pm 1$$
From the options, $m=1$.