1. **State the problem:** Solve the differential equation $$\sec^2 x \tan y \, dx + \sec^2 y \tan x \, dy = 0.$$\n\n2. **Rewrite the equation:** Let $$M(x,y) = \sec^2 x \tan y$$ and $$N(x,y) = \sec^2 y \tan x.$$ The equation is $$M \, dx + N \, dy = 0.$$\n\n3. **Check if the equation is exact:** Compute partial derivatives:\n$$\frac{\partial M}{\partial y} = \sec^2 x \sec^2 y,$$\n$$\frac{\partial N}{\partial x} = \sec^2 y \sec^2 x.$$\nSince $$\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x},$$ the equation is exact.\n\n4. **Find the potential function $$\psi(x,y)$$:** Integrate $$M$$ with respect to $$x$$:\n$$\psi(x,y) = \int \sec^2 x \tan y \, dx = \tan y \int \sec^2 x \, dx = \tan y \tan x + h(y),$$ where $$h(y)$$ is a function of $$y$$ only.\n\n5. **Differentiate $$\psi$$ with respect to $$y$$ and equate to $$N$$:**\n$$\frac{\partial \psi}{\partial y} = \sec^2 y \tan x + h'(y) = N = \sec^2 y \tan x.$$\nThis implies $$h'(y) = 0,$$ so $$h(y)$$ is a constant.\n\n6. **Write the implicit solution:**\n$$\psi(x,y) = \tan y \tan x = C,$$ where $$C$$ is a constant.\n\n**Final answer:** $$\tan x \tan y = C.$$