Proving $σ_{ji,j} + f_i = ρ(Dv_i/Dt) \implies (σ_{ij} - ρv_iv_j) + f_i = \partial(ρv_i)/ \partial t$ Exercise :

Prove that
  $$σ_{ji,j} + f_i = ρ(Dv_i/Dt) \implies (σ_{ij} - ρv_iv_j)_{,j} + f_i = \partial(ρv_i)/ \partial t$$
  where $σ_{ij} = σ_{ji}$ is the stress tensor.

Attempt :
It is : 
$$\frac{Dv_i}{Dt} = \frac{\partial v_i}{\partial t} + v_j\frac{\partial v_i}{\partial x_j}$$
Thus, we yield :
$$σ_{ji,j} + f_i = ρ\bigg(\frac{\partial v_i}{\partial t} + v_j\frac{\partial v_i}{\partial x_j}\bigg) \Rightarrow σ_{ji,j} + f_i -ρv_j\frac{\partial v_i}{\partial x_j} = ρ\frac{\partial v_i}{\partial t}$$
$$\implies$$
$$\dots$$
How does one continue and use the Continuity Equation to yield the desired expression ? It starts to form since the inded $,j$ means the differential with respect to $j$ coordinate variable and we have aspects of that there, but I cannot finalise it.

Try this:
$$\begin{align}
\rho \frac{\partial v_i}{\partial t}
&= \frac{\partial}{\partial t} \left( \rho v_i \right) - \frac{\partial \rho}{\partial t} v_i \\
&= \frac{\partial}{\partial t} \left( \rho v_i \right) - \left( \nabla \cdot (\rho v) \right) v_i \\
&= \frac{\partial}{\partial t} \left( \rho v_i \right) - \left( \partial_j (\rho v_j) \right) v_i \\
&= \frac{\partial}{\partial t} \left( \rho v_i \right) - \partial_j (\rho v_j v_i) + \rho v_j \partial_j v_i\\
\end{align}$$
