Taking the curl of advective part of navier-stokes equation to get vorticity in index notation I need to take the curl of
$\frac{\partial u_i}{\partial t} = -\frac{1}{\rho}\frac{\partial p}{\partial x_i}-\frac{\partial}{\partial x_j}u_i u_j+\nu\frac{\partial^2 u_i}{\partial x_i^2}$
to get
$\frac{\partial \omega_p}{\partial t} =\omega_k\frac{\partial u_p}{\partial x_k} - u_k\frac{\partial \omega_p}{\partial x_k} + \nu\frac{\partial^2 \omega_p}{\partial x_k^2}$
I managed to get all the terms except of
$\omega_k\frac{\partial u_p}{\partial x_k} - u_k\frac{\partial \omega_p}{\partial x_k}$
When I opened $\frac{\partial}{\partial x_j}u_i u_j$ up I reached:
$\epsilon_{ijk}\partial_j u_p\partial_p u_k+u_p\partial_p\omega_i$
so I really only need to figure out how to get $\omega_k\frac{\partial u_p}{\partial x_k}$ from $\epsilon_{ijk}\partial_j u_p\partial_p u_k$
Thanks
 A: Presumably, you have incompressible flow with $\nabla \cdot \mathbf{u}=0$. Otherwise, you will not get the form you are looking for.
With vorticity $\mathbf{\omega} := \nabla \times \mathbf{u}$, the result you are trying to find is
$$\tag{1}\nabla \times \nabla \cdot (\mathbf{u} \mathbf{u})=  -\mathbf{\omega}\cdot \nabla\mathbf{u}+ \mathbf{u} \cdot \nabla \mathbf{\omega},  $$
and it is difficult to derive directly without knowing some relationships that are not immediately obvious.
Proceeding as you did, using Cartesian components and the Einstein summation convention, we have
$$\nabla \times \nabla \cdot (\mathbf{u} \mathbf{u}) \cdot \mathbf{e}_i = \epsilon_{ijk}\partial_j\partial_p(u_ku_p) = \underbrace{\epsilon_{ijk}\partial_ju_p\partial_pu_k}_{A} + \underbrace{\epsilon_{ijk}\partial_ju_k(\partial_pu_p)}_{B}$$
Notice that term $B$ is not $u_i\partial_p\omega _i$ as you wrote, but rather
$$B = \epsilon_{ijk}\partial_ju_k(\partial_pu_p)= \omega_i(\partial _pu_p),
$$
which is the $i-$th component of $(\nabla\cdot \mathbf{u})\mathbf{\omega}$.  For incompressible flow, we therefore have $B=0$.
Term $A$ is the $i-$th component of $\nabla \times (\mathbf{u} \cdot \nabla \mathbf{u}$). To get the desired form on the RHS of (1), it is extremely helpful to know the identity
$$\mathbf{u} \cdot \nabla \mathbf{u}= \nabla (\frac{1}{2} \mathbf{u}\cdot \mathbf{u}) - \mathbf{u}\times \omega$$ as well as the fact that the curl of a gradient is zero.  Hence, $\nabla \times \nabla (\frac{1}{2} \mathbf{u}\cdot \mathbf{u})= 0$ and the term $A$ is the $i-$th component of $-\nabla \times (\mathbf{u} \times \mathbf{\omega})$, that is
$$A = -\epsilon_{ijk}\partial_j\epsilon_{kpq}u_p\omega_q =  -\epsilon_{ijk}\epsilon_{kpq}\partial_j(u_p\omega_q)$$
The product of the Levi-Civita symbols contracted on one index can be written in terms of Kronecker deltas as
$$\epsilon_{ijk}\epsilon_{kpq}= \delta_{ip}\delta_{jq} - \delta_{iq} \delta_{jp},$$
and, hence,
$$\tag{2}A = -\delta_{ip}\delta_{jq}\partial_j(u_p\omega_q)+ \delta_{iq} \delta_{jp}\partial_j(u_p\omega_q)= -\partial_j(u_i \omega_j)+ \partial _j (u_j \omega_i)\\= -\omega_j\partial_ju_i-u_i\partial_j\omega_j + u_j\partial_j\omega_i+ \omega_i\partial_ju_j$$
Since the divergence of the curl of a vector field is zero we have $\nabla\cdot \mathbf{\omega} = \nabla \cdot \nabla \times \mathbf{u}=0$ and for the second term on the RHS of (2) we have $u_i\partial_j \omega_j=0$
Since $\partial_ju_j = 0$ for incompressible flow, it follows that
$$\tag{3}A +B = A=-\omega_j \partial_j u_i +u_j\partial_j \omega_i + (\partial_ju_j)\omega_i =  -\omega_j \partial_j u_i +u_j\partial_j \omega_i, $$
which is (1) in component form.
