$$ \frac{∂u}{∂t}+u\frac{∂u}{∂x}+v\frac{∂u}{∂y}+w\frac{∂u}{∂z}=-\frac{1}{ρ}\frac{∂P}{∂x}+gx+\nu\left(\frac{∂^2u}{∂x^2}+\frac{∂^2u}{∂y^2}+\frac{∂^2u}{∂z^2}\right) $$
Why would someone use a form of this equation where both sides are not divided by the density? Are there any cases where the above equation cannot be used because the equation has been divided by the density?
Let's take the general conservative compressible form of the mass continuity \eqref{1} and momentum equations \eqref{2}
$\frac{\partial \rho}{\partial t} + \sum\limits_{j \in \mathcal{D}} \frac{\partial (\rho u_j )}{\partial x_j }=0 \tag{1}\label{1}$
$\frac{\partial (\rho u_i )}{\partial t} + \sum\limits_{j \in \mathcal{D}}\frac{\partial (\rho u_i u_j )}{\partial x_j} = \sum\limits_{j \in \mathcal{D}} \frac{\partial \sigma_{ij}}{\partial x_j } + \rho g_i \tag{2}\label{2}$
where $\mathcal{D} = \{ x, y, z \}$ and assume a Newtonian fluid with the Stokes' hypothesis $\sigma_{ij} = - p \delta_{ij} + 2 \mu S_{ij} \tag{3}\label{3} - \frac{2}{3} \mu \sum\limits_{k \in \mathcal{D}} S_{kk} \delta_{ij}$
where the rate of strain tensor is given by
$$S_{ij} := \frac{1}{2} \left( \frac{\partial u_i}{\partial x_j} + \frac{\partial u_j}{\partial x_i} \right). \tag{4}\label{4}$$
Now let's go through what simplification have been made to arrive at your form.
First insert equation \eqref{3} and \eqref{4} into \eqref{2} and apply the chain rule
$$\underbrace{u_i \frac{\partial \rho}{\partial t} + u_i \sum\limits_{j \in \mathcal{D}}\frac{\partial (\rho u_j)}{\partial x_j}}_{u_i \cdot (1) = 0} + \rho \frac{\partial u_i}{\partial t} + \sum\limits_{j \in \mathcal{D}} (\rho u_j) \frac{\partial u_i}{\partial x_j} = - \frac{\partial p}{\partial x_i } + \sum\limits_{j \in \mathcal{D}} \frac{\partial}{\partial x_j } \Big[ \mu \big( \frac{\partial u_i}{\partial x_j} + \underbrace{ \frac{\partial u_j}{\partial x_i} }_{II}\big) \underbrace{ - \frac{2}{3} \mu \sum\limits_{k \in \mathcal{D}} \frac{\partial u_k}{\partial x_k} \delta_{ij} }_{I} \Big] + \rho g_i.$$
The first term cancels out due to continuity equation \eqref{1} and we are left with the non-conservative form of the momentum equation. Your equation looks fairly similar but we instantly see that the shear viscosity $\mu$ is in your formula replaced by the kinematic viscosity $\nu := \frac{\mu}{\rho}$ and it stands outside the derivative - it is assumed constant.
Furthermore the term $- \frac{2 \mu}{3} \sum\limits_{k \in \mathcal{D}} \frac{\partial u_k}{\partial x_k}$ corresponding to the volumetric dilatation is not there. This means a fluid parcel is not compressed along its way on a stream line
$$\frac{D \rho}{D t} = \frac{\partial \rho}{\partial t} + \sum\limits_{j \in \mathcal{D}} u_j \frac{\partial \rho}{\partial x_j} = 0,$$
something termed incompressible flow (it could also be slightly more strict and be an incompressible fluid). With equation \eqref{1} we can find - applying the chain rule - that in such a case the velocity field is divergence free
$$\sum\limits_{j \in \mathcal{D}} \frac{\partial u_j}{\partial x_j} = 0.$$
This is the reason why the two terms I and II are equivalent to zero and not present in your equation.
Summing up, these equations are only valid for incompressible flow of a Newtonian fluid with a constant viscosity.
