已知 $\sin \alpha+2 \cos \alpha=0$, 则 $\frac{\cos 2 \alpha}{\sin 2 \alpha-\cos ^{2} \alpha}=$ ( )
由题意知 $\sin \alpha+2 \cos \alpha=0$, 即 $\sin \alpha=-2 \cos \alpha$, 可得 $\tan \alpha=-2$,

又由 $\frac{\cos 2 \alpha}{\sin 2 \alpha-\cos ^{2} \alpha}=\frac{\cos ^{2} \alpha-\sin ^{2} \alpha}{2 \sin \alpha \cos \alpha-\cos ^{2} \alpha}=\frac{1-\tan ^{2} \alpha}{2 \tan \alpha-1}=\frac{3}{5}$, 故选 D.