e^2xe ^{- 3y}dx + e^2ye ^{- 3x}dy = 0

Rearringing the terms we get:

e^{2x + 3x}dx = - e^{2y + 3y}dy

⇒e^5xdx = - e^5ydy

Integrating both the sides we get:

⇒e^5x + e^5y = 5c’ = c

Ans: e^5x + e^5y = c