Let $\partial u_i/\partial x_i=0$ then given that
$$\sigma_{ij} = -p\delta_{ij}+\mu\left(\frac{\partial u_i}{\partial x_j}+\frac{\partial u_j}{\partial x_i}\right)$$
Show that
$$\frac{\partial \sigma_{ij}}{\partial x_j} = -(\nabla p)_i+\mu\nabla^2u_i$$
So the solution is:
\begin{align} \frac{\partial \sigma_{ij}}{\partial x_j} &= \frac{\partial}{\partial x_j} \left(-p\delta_{ij}+\mu\left(\frac{\partial u_i}{\partial x_j}+\frac{\partial u_j}{\partial x_i}\right)\right) \\ &= - \frac{\partial p}{\partial x_j} + \mu\left(\frac{\partial^2 u_i}{\partial x_j\partial x_j}+\frac{\partial^2 u_j}{\partial x_j\partial x_i}\right) \\ &= - \frac{\partial p}{\partial x_j} + \mu\frac{\partial^2 u_i}{\partial x_j\partial x_j} \tag{since $\partial u_i/\partial x_i=0$} \\ &= - (\nabla p)_i + \mu\nabla^2 u_i \end{align}
But why on the first to the second line have we suddenly removed $\delta_{ij}$ (which is the kronecker delta)?
$\partial_j (p \delta_{ij}) = \partial_j p \delta_{ij} + p \partial_j \delta_{ij} $
but $\delta_{ij}$ is a constant, so its derivative is zero, and we are left with
$\partial_j p \delta_{ij} = \partial_i p$