= Solution
The first argument mixes a material <fluid element> with a fixed spatial volume. For a <material volume element> following the fluid, its volume is generally time dependent:
$$
\frac D{Dt}\delta V=(\nabla\cdot\mathbf u)\delta V.
$$
Therefore the <material derivative> cannot be taken outside the product with $\delta V$ as in the proposed cancellation. Instead, <mass conservation> states
$$
\frac D{Dt}(\rho\delta V)=0,\qquad
\frac{D\rho}{Dt}=-\rho\nabla\cdot\mathbf u.
$$
The <momentum balance for a material fluid element>, using <Newton's second law>, is consequently
$$
\frac D{Dt}(\rho\delta V\,\mathbf u)
=(\rho\delta V)\frac{D\mathbf u}{Dt}
=-\nabla p\,\delta V,
\qquad
\boxed{\rho\frac{D\mathbf u}{Dt}=-\nabla p.}
$$
For a fixed <control volume>, fluid crosses its boundary and carries <momentum>; a <momentum flux> term must be included. Its local conservative form is $\partial_t(\rho\mathbf u)+\nabla\cdot(\rho\mathbf u\otimes\mathbf u)=-\nabla p$, equivalent to the displayed <Euler equations> after using the <continuity equation>. The erroneous equation omits the term $\rho\mathbf u\nabla\cdot\mathbf u$ needed to convert a material derivative of momentum density into this conservative form.
For the second task, take a homogeneous fluid with constant <mass density> $\rho$, prescribed gravitational <acceleration> $\mathbf g$, <incompressible flow> and <inviscid flow>. Its equations are
$$
\frac{D\mathbf u}{Dt}=-\nabla\left(\frac p\rho\right)+\mathbf g,\qquad\nabla\cdot\mathbf u=0.
$$
Under the replacement $\rho\mapsto\widetilde\rho=C\rho$, the same <velocity> solves these equations if $p\mapsto\widetilde p=Cp$, up to a spatially constant pressure offset. This is <density scaling invariance of incompressible Euler flow>:
$$
\boxed{\text{The velocity field alone cannot determine the uniform fluid density.}}
$$
The comparison assumes the same geometry and compatible velocity or gravitational <free surface> conditions; an independently specified nonuniform boundary <pressure> would supply additional information. The <gravitational field> is taken as prescribed. If the fluid's own <self-gravity> must be included, its field changes with <mass density>, so the invariance need not hold. A single freely moving particle's density-independent gravitational acceleration does not justify neglecting that distinction.
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