${\sec ^2}x\tan ydx + {\sec ^2}y\tan xdy = 0$

.: We have, ${\sec ^2}x\tan ydx + {\sec ^2}y\tan xdy = 0$

$\Rightarrow \cfrac{{{{\sec }^2}x}}{{\tan x}}dx + \cfrac{{{{\sec }^2}y}}{{\tan y}}dy = 0$
…(1)
Integrating (1) both sides,

we get
$\int {\cfrac{{{{\sec }^2}x}}{{\tan x}}} dx + \int {\cfrac{{{{\sec }^2}y}}{{\tan y}}} dy = 0$

$\Rightarrow \log |\tan x| + \log |\tan y| = \log C$

$\Rightarrow \log |\tan x\tan y| = \log C \Rightarrow \tan x\tan y = C$

which is required solution.