The square of the norm of the velocity field is
$$|\mathbf{u}|^2=u_x^2+u_y^2+u_z^2.$$
Taking the partial derivative with respect to time we get
$$\frac{\partial}{\partial t}|\mathbf{u}|^2=2u_x\frac{\partial u_x}{\partial t}+2u_y\frac{\partial u_y}{\partial t}+2u_z\frac{\partial u_z}{\partial t}= 2\mathbf{u} \cdot\frac{\partial \mathbf{u}} {\partial t}.$$
Hence,
$$\mathbf{u} \cdot\frac{\partial \mathbf{u}} {\partial t}=\frac1{2}\frac{\partial}{\partial t}|\mathbf{u}|^2$$
The mass of a fluid within a volume $V$ is,
$$\int_V \rho \ dV$$
Where, $\rho$ is the density. Similarly, the momentum is,
$$\int_V \rho \cdot \vec v \ dV$$
Where $\vec v$ is the fluid velocity at a point inside the volume.
The mass within the volume changes with time if fluid flows in and out of the bounding surface $S$ of the volume.
$${{d} \over {dt}} \cdot \int_V \rho \ dV=-\int_S \rho \cdot \vec v \cdot \vec n \ dS$$
Where we take the dot product when appropriate and $\vec n$ is the surface normal. Similarly,
$${{d} \over {dt}} \cdot \int_V \rho \cdot \vec v \ dV=-\int_S \rho \cdot \vec v \ ( \vec v \cdot \vec n) \ dS$$
Using Newton's Second Law, the rate of change of momentum of a fixed portion of material is equal to the net force acting on that material. If we assume the force is due to pressure $p$ at the surface of $V$, we can say,
$$(1) \quad {{d} \over {dt}} \cdot \int_V \rho \cdot \vec v \ dV=-\int_S \rho \cdot \vec v \ ( \vec v \cdot \vec n) +p \cdot \vec n \ dS$$
Now, we can use the fact that,
$$\int_V \nabla f \ dV=\int_S \vec n \cdot f \ dS$$
along with the fact,
$$\int_V \nabla \cdot \vec B \ dV=\int_S \vec n \cdot \vec B \ dS$$
and rewrite $(1)$ as,
$$(2) \quad {{d} \over {dt}} \cdot \int_V \rho \cdot \vec v \ dV=-\int_V \nabla \cdot (\rho \cdot \vec v) \cdot \vec v +\nabla p\ dV$$
Moving terms, we get,
$$(3) \quad \int_V {{d} \over {dt}} \ (\rho \cdot \vec v)+\nabla \cdot (\rho \cdot \vec v) \cdot \vec v +\nabla p\ dV=0$$
Which can be simplified to,
$$(4) \quad {{d} \over {dt}} \ (\rho \cdot \vec v)+\nabla \cdot (\rho \cdot \vec v) \cdot \vec v +\nabla p=0$$
Which finally becomes,
$$(5) \quad {{D} \over {Dt}} \ (\rho \cdot \vec v)+\nabla p=0$$
Which is the Euler Equation for inviscid flow.
Your Method
I commented on your method
The scanty memory that I have tells me that I must assume that the fluid is at rest and hence consider that the only forces exerted on the fluid are pressure and gravitational pull and that the force due to pressure is given by
$$ \nabla P dxdydz $$
This is completely correct
and the force due to possible outward acceleration $\overrightarrow{a}$ is $$\overrightarrow{F} $$
Ad-hoc, but correct
so that the total outward force is
$$ \nabla P dx \cdot dy \cdot dz + \overrightarrow{F} $$
Yes, that is true
And by newton III, that is $F_{AB}=-F_{BA}$ , I think that this force is reacted back by the weight of the liquid $ g dm = g \rho dx \cdot dy \cdot dz$ where $dm$ is elemental mass in the fluid and hence I can write $$ \nabla P dx \cdot dy \cdot dz + \overrightarrow{F} = - g \rho dxdydz \overrightarrow{ k }$$
Use Newton's Second Law and I think this works
Best Answer
Half a year had past and it's ripe to answer the question :-)
It is fairly easy. First, since Archimedes’ Principle is essentially about hydrostatics, then $\boldsymbol u = 0$, i.e. nothing moves. Thus your equation turns into
$$\nabla p = - \rho \boldsymbol g$$
Then you have to integrate pressure over the whole body's surface to find the force exerted on the body by the fluid:
$$\boldsymbol F = \int_S p \cdot d \boldsymbol s = \int_V \nabla p \, dV = - \rho \boldsymbol g \int_V dV = - \rho V \boldsymbol g$$
$V$ is the volume of the body, $S$ is its surface. That's all. I've used Gauss theorem here:
$$\int_V \left(\nabla\cdot\mathbf{F}\right)dV=\int_S(\mathbf{F}\cdot\mathbf{n}) \, dS$$
However as you see I've integrated $\nabla p$ inside the body, just like there was fluid and no body. If you are a mathematician, I think you may say that given $p(\boldsymbol x)$ satisfying the hydrostatic equation outside the body, there is a unique smooth continuation of $p$ inside the body, satisfying the same equation.
But I'm rather a physician and I would say, that we can substitute the body by a bubble full of fluid and the forces acting on that bubble from the outside fluid won't change. That's just because these forces depend only on the shape of the body.
UPD Oops, I've missed the minus sign, I've corrected it --- the force exerted on the body is opposite to $\boldsymbol g$.