What is really an exact differential and how does it relate to conservative fields

If there is a differential form $$A(x,y,z) dx + B(x,y,z) dy + C(x,y,z) dz$$ where there exists some function $$\psi(x,y,z)$$

Let $$\psi = \psi (x,y,z)$$

Then the total differential is $$d \psi = \left(\frac{\partial \psi}{\partial x} \right) dx + \left(\frac{\partial \psi}{\partial y} \right) dy + \left(\frac{\partial \psi}{\partial z} \right) dz$$

If $$A = \left(\frac{\partial \psi}{\partial x} \right)$$ , $$B = \left(\frac{\partial \psi}{\partial y} \right)$$ , $$C = \left(\frac{\partial \psi}{\partial z} \right)$$

Then the form $$A(x,y,z) dx + B(x,y,z) dy + C(x,y,z) dz$$ is said to be 'exact'.

First, what does being 'exact' mean?

Also what does this have to do with independence of path if we integrate $$\int_i^f d\psi = \psi(f) - \psi(i)$$

Why does an exact differential imply a conservative field?

• $A(x,y,z)\, dx + B(x,y,z)\, dy + C(x,y,z) \,dz$ is not an ‘equation’, as you call it (there is no unknown to be determined) but a differential form (more precisely a differential $1$-form). A differential $1$-form which is the differential of a function is said to be exact. – Bernard Feb 6 at 18:35