In deriving Euler's equations for fluid mechanics, in particular
$$f=\rho \partial_t v +\rho v\cdot \nabla v$$
for some body force $f$ (e.g Landau & Lifschitz 2.3) one assumes the continuity equation $\partial_t\rho=-\nabla\cdot\left(\rho v\right)$ and Newton's second law in the form $F=ma$ so $$f=\rho D_t v$$ where $D_t=\partial_t+v\cdot \nabla$ is the total time derivative (e.g Landau & Lifschitz 2.1).
If one instead uses $F=\dot p$ so that $$f = D_t\left(\rho v\right)$$ then in order to get to get our initial equation back one must assume $D_t \rho=0$ but of course $D_t \rho=\partial_t\rho+v\cdot \nabla\rho=\nabla\cdot\rho$ and so we have assumed the flow is incompressible.
Landau implies that he has not assumed incompressibility at this point so how is it that one can choose the first form for Newton's second law without loss of generality?
(If your answer would appeal to stress-energy tensors or Navier-Stokes etc., if possible please show how the assumption isn't implicitly made.)
To be more clear, as I have shown above, if you say that $f = D_t\left(\rho v\right)$ and also that $f=\rho \partial_t v +\rho v\cdot \nabla v$ then it follows that the flow is incompressible. I don't think that this does imply incompressibility so why is the assumption false?
Best Answer
The main problem with continuum equations is that it is just a "lucky guess" of macroscopic dynamics otherwise governed by microscopic ones. You have encountered the moment where this "guessing" isn't entirely consistent.
Before delving into the technical derivation, the non-$D_t \rho$ in the Newton's equation is just a consequence of the fact that the actual force acts on microscopic particles whose mass doesn't change. Hence, only the macroscopic velocity of the fluid is affected by the force.
Let us define a one particle phase-space distribution function $f(\vec{x},\vec{p},t)$. We have the Boltzmann equation derived straightforward from classical mechanics (this is actually the only point where $\vec{F}=\dot{\vec{p}}$ enters) $$\frac{df}{dt} = \partial_t f + \{f,H\} = \partial_t f + \frac{\vec p \cdot \nabla_x f}{m} + \vec{F} \cdot \nabla_p f$$ Without colliding with other particles we would just have $df/dt = 0$ but with interaction we can describe the ensemble as N copies of a one-particle distribution with a collision term $$\partial_t f + \frac{\vec p \cdot \nabla_x f}{m} + \vec{F} \cdot \nabla_p f = \delta_t f|_c$$ We can now integrate the whole equation in momentum space $\int d^3p$ to get the continuity equation $$\partial_t \rho + \nabla \cdot (\rho \vec{v}) = \delta_t \rho|_c$$ With naturally $\rho \equiv m \int f d^3p$ and $\vec{v} = <\vec{p}>/\rho = \int \vec{p} f d^3 p/\rho$. The term $\delta_t \rho|_c$ is non-zero for example in chemical reactions. Integrating again the Boltzmann equation, this time like $\int \vec{p} ... d^3 p$, we get $$\partial_t(\rho v^j) + \sum \partial_{x_i} T_{ij} - \frac{\rho}{m} F^j = \delta_t p^j|_c$$ Where again $\delta_t p|_c$ is zero for momentum conserving collisions and no chemical reactions. The new symbol is $T_{ij} = <p_i p_j>/m = \int p_i p_j f d^3p/m$ and you can identify in it $ \rho v^i v^j$, isotropic pressure ($<\sum p_i^2>/3\rho$) and viscous stress ($(<p_i p_j>-<p_i><p_j>)/\rho$). For simplicity, let us now drop the viscous stress and using the continuum equation we can get $$\frac{\rho}{m} \vec{F} = \rho D_t \vec{v} + \nabla P ,$$ or a much more "Newton law" form $$\vec{F} = m D_t \vec{v} + \frac{m}{\rho} \nabla P.$$ So the Newton's law in continuum is derivable only as an intuition of how the sum of forces on individual particles will act on the average properties of the ensemble. It all make sense when you consider that $\vec{f}$ is not exactly the force but force time the number of particles per volume.
However, notice that this is the equation for mean velocity, but the force affects also higher momenta($\int \vec{p} \vec{p}...d^3p$) of the momentum equation and this is important for example in plasma physics.
In non-extreme fluids the higher momentum equations are usually neglected as the higher momenta of $\delta_t f|_c$ do not vanish so easily and are actually very difficult to derive in a non-perturbative regime (which is the case of a tightly-packed fluid rather than a loose plasma or gas). As a consequence, observed macroscopic properties of the fluid such as incompressibility are plugged in as extra conditions to get a practical model.
My derivations are very brief, for details see e.g. http://www.astronomy.ohio-state.edu/~dhw/A825/notes2.pdf (with a different notation).
EDIT: Combining my memory using the phase-space notation with the notation of the linked text, I have done some possibly quite confusing typos. Now the equations should be all correct. I also added a commentary to the much clearer form of the final "Newton's law".