As we know, the definition of material derivative of $\varphi$ is: $\frac{D\varphi}{Dt}\equiv \frac{\partial \varphi}{\partial t} + \mathbf{u}\cdot\nabla\varphi$. And the physical meaning of material derivative is the change of $\varphi$ of a fluid parcel moving with the stream.
And it is easy to find that the continuity equation (${\displaystyle {\frac {\partial \rho }{\partial t}}+\nabla \cdot \mathbf {(\rho u)} =0}$) means the material derivative of the density is 0, which also means that every fluid parcel's density is constant along their way.
This cannot be understood intuitively easily, why the continuity equation means constant densities of each fluid parcel?
I think this SE question is relavent, but I still look forward to some answers based on my question's view.
Best Answer
Your statement is not true in general. On the one hand you have: $$ \frac{D\rho}{dt} = \partial_t \rho + \vec u \cdot \nabla \rho, $$ On the other hand the continuity equations (which implies conservation of total mass) gives: $$ 0 = \partial_t \rho + \nabla \cdot (\vec u \rho) = \partial_t \rho + \rho \nabla \cdot \vec u + \vec u \cdot \nabla \rho. $$
So the continuity equation implies that the material derivative is zero only for a flow whose velocity field is divergence free, and this exactly characterizes incompressible flow. (Although usually it is derived the other way around, you compare the material derivative and the continuity equation and see that for incompressibility, $\nabla \cdot \vec u$ has to vanish).
TL;DR: The continuity equation for the mass density only implies conservation of mass not conservation of density for fluid parcels. Incompressibility is characterized by $D_t \rho = 0$, and it can be shown to be equivalent to the continuity equation plus $\nabla \cdot \vec u = 0$.