Tham khảo thôi nhé!Đặt $x=2y\tan a\Rightarrow \frac{x^2-(x-4y)^2}{x^2+4y^2}=\frac{\tan^2a-(\tan a-2)^2}{1+\tan^2a}=4(\tan a-1).\cos^2a$
$=2\sin 2a-2(1+\cos 2a)=2(\sin 2a-\cos 2a)-2$
$=2\sqrt{2}\sin(2a-\frac{\pi}{2})-2\in \left[ {}-2\sqrt{2} -2;2\sqrt{2}-2\right]$