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 the  value of ‘m’ is

Angle bisectors of xy=0 are y=x and y=-x

  $$\begin{gathered} Given pair of lines m{y}^2+(1-m^2)xy-m{x}^2=0 \\ \Rightarrow my(y-mx)+x(y-mx)=0 \end{gathered}$$
i.e.,  $(y-mx)\times (my+x)=0$

So, the lines are  $y=mx$ and  $y=-\frac{1}{m}x$. 

compare with y=x or y=-x then we get

  $m=1$ or  $-1$.