10 A. Let $z=r(\cos \theta+i \sin \theta)$, then
$$
-\frac{1}{z^{2}}=-\frac{1}{r^{2}}(\cos 2 \theta-i \sin 2 \theta)=\frac{1}{r^{2}}(\cos (\pi-2 \theta)+i \sin (\pi-2 \theta)) .
$$
$0<\pi-2 \theta<\frac{\pi}{2}$, so $-\frac{1}{z^{2}}$ corresponds to a point in the first quadrant.
10 Let the complex number $z$ satisfy the condition $\frac{\pi}{4}<\arg z<\frac{\pi}{2}$, then the point corresponding to $-\frac{1}{z^{2}}$ is located in
A. the first quadrant
B. the second quadrant
C. the third quadrant
D. the fourth quadrant