$x \cos^2 y \, dx + \tan y \, dy = 0$
$x \, dx + \dfrac{\tan y \, dy}{\cos^2 y} = 0$
$x \, dx + \tan y \sec^2 y \, dy = 0$
$\displaystyle \int x \, dx + \int \tan y \sec^2 y \, dy = 0$
$\frac{1}{2}x^2 + \frac{1}{2}\tan^2 y = \frac{1}{2}c^2$
$x^2 + \tan^2 y = c^2$ answer