Remember implicit differentiation?

xyz = e^z

Take the partial "differential" of both sides with respect to x, that is take y to be a constant for a moment:

yz(dx) + xy(dz) = e^z(dz)

Then

yz + xy(dz/dx) = e^z(dz/dx) <-- These are now partial derivatives.

dz/dx = yz/(e^z - xy) = yz/(xyz - xy)

Similarly:

xz(dy) + xy(dz) = e^z(dz)

xz + xy(dz/dy) = e^z(dz/dy)

dz/dy = xz/(e^z - xy) = xz/(xyz - xy)

Thus:

dz = [yz/(xyz - xy)]dx + [xz/(xyz - xy)]dy <-- The "d's" are now full differentials. Sorry about the notational confusion!

-Dan