total differential
If a quantity F depends on two inputs x and y, and you nudge both by tiny amounts dx and dy, how much does F change? The answer, to first order, is the total differential: you add up the change caused by moving in x and the change caused by moving in y. It is the two-variable version of the familiar one-variable statement dF = F'(x) dx.
Written out, the total differential of F(x, y) is dF = (partial F / partial x) dx + (partial F / partial y) dy. The coefficient of dx is the rate of change of F as you slide in the x-direction holding y fixed (a partial derivative); the coefficient of dy is the rate as you slide in the y-direction. So a combination M dx + N dy is a genuine total differential of some F precisely when you can find an F whose two partial derivatives are M and N.
This is the engine behind exact equations. The equation M dx + N dy = 0 is asking: is the left side a total differential dF? If yes, the equation reads dF = 0, meaning F stays constant, and F(x, y) = C solves it. If no, the left side is not the differential of any single function, and you cannot integrate it so cleanly — that is the case an integrating factor tries to fix.
If F(x, y) = x^2 y, then dF = 2xy dx + x^2 dy. So the combination 2xy dx + x^2 dy is a total differential, and 2xy dx + x^2 dy = 0 is an exact equation with solution x^2 y = C.
dF packages both partial derivatives into one differential — the heart of exactness.
Not every M dx + N dy is a total differential. The combination y dx (with N = 0) is not, because no F has partial F / partial x = y and partial F / partial y = 0 simultaneously.