Let suppose that we work in a 2D cartesian coordinates. what will be x and y components of $\nabla.\left[-p I+\mu \left(\nabla \text{u}+(\nabla \text{u})^T\right)-\frac{2}{3} \mu (\nabla.\text{u}) I \right]$
$p=p(x,y)$ is pressure, $\mu$ is viscosity and constant, $I$ is identity tensor and $\text{u}=u \hat i +v \hat j$ is velocity field.
Rule of thumb: the divergence of a matrix (a second order tensor) can be defined by the trace of the gradient of a tensor field.
For a $2\times 2$ matrix, the divergence is actually the divergence of each row: $$ \nabla\cdot A = \begin{pmatrix}\nabla \cdot A_{1i} \\ \nabla \cdot A_{2i} \end{pmatrix} = \begin{pmatrix}\partial_{x} A_{11}+\partial_y A_{12} \\ \partial_{x} A_{21} + \partial_y A_{22} \end{pmatrix}. $$
Pressure term $pI$: $$ \nabla\cdot (pI) = \begin{pmatrix}\partial_{x} p \\ \partial_{y} p\end{pmatrix} = \nabla p. $$
The first term in the stress tensor is the first order strain tensor (by some factor of constants) $\nabla \mathbf{u}+(\nabla \mathbf{u})^T$: firstly when $\mathbf{u} = (u_1, u_2) = u_1 \mathbf{i} + u_2 \mathbf{j}$, $$ \nabla \mathbf{u} = \begin{pmatrix}\partial_{x} \mathbf{u}_1 & \partial_y\mathbf{u}_1 \\ \partial_{x} \mathbf{u}_2 & \partial_y\mathbf{u}_2 \end{pmatrix}, $$ and taking the divergence: $$ \nabla\cdot\nabla \mathbf{u} = \begin{pmatrix}\partial_{xx} \mathbf{u}_1 & \partial_{yy}\mathbf{u}_1 \\ \partial_{xx} \mathbf{u}_2 & \partial_{yy}\mathbf{u}_2 \end{pmatrix}. $$ $\nabla \cdot (\nabla \mathbf{u})^T$ is left for you.
For the second term in the stress tensor $(\nabla \cdot \mathbf{u}) I$: $$ (\nabla \cdot \mathbf{u}) I = \begin{pmatrix}\nabla \cdot \mathbf{u} & 0 \\ 0 & \nabla \cdot \mathbf{u}\end{pmatrix}, $$ and taking the divergence: $$ \nabla\cdot \Big((\nabla \cdot \mathbf{u}) I\Big) = \begin{pmatrix} \partial_x (\nabla \cdot \mathbf{u} ) \\ \partial_y (\nabla \cdot \mathbf{u})\end{pmatrix} = \nabla (\nabla \cdot \mathbf{u}). $$