$\begin{aligned} & \sec ^{2} x \tan y d x+\sec ^{2} y \tan x d y=0 \\ \therefore & \sec ^{2} x \tan y d x=-\sec ^{2} y \tan x d y \\ & \frac{\sec ^{2} x}{\tan x} d x=-\frac{\sec ^{2} y}{\tan y} d y \Rightarrow \int \frac{\sec ^{2} x}{\tan x} d x=-\int \frac{\sec ^{2} y}{\tan y} d y \\ & \log (\tan x)=-\log (\tan y)+\log c \\ \therefore & \log (\tan x \tan y)=\log c \\ & \tan x \tan y=c \end{aligned}$