Given: $x^2+4xy-y^2=0$

This is the equation of pair of straight lines so comparing it with $a{x}^2+2hxy+b{y}^2=0$ we get, 

$a=1, b=-1, h=2$

As we know the equation of the bisector is given by, $\frac{x^2-y^2}{a-b}=\frac{xy}{h}$

$\Rightarrow \frac{x^2-y^2}{1-\left(-1\right)}=\frac{xy}{2}$

$\Rightarrow x^2-xy-y^2=0$

As $y=mx$ is one of the bisector

$\Rightarrow x^2-x\left(mx\right)-{\left(mx\right)}^2=0$

$\Rightarrow x^2-m{x}^2-m^2{x}^2=0$

$\Rightarrow m^2+m-1=0$

$\Rightarrow m=\frac{-1\pm \sqrt{5}}{2}$

$\Rightarrow 2m=-1\pm \sqrt{5}$