Before proceeding, let us prove the identity $\nabla\cdot(\vec{a}\vec{b}) = \vec{b}\cdot\nabla\vec{a} + \vec{a}\nabla\cdot\vec{b}$. The derivation becomes easy when we use the cartesian tensor form, \begin{equation} \frac{\partial}{\partial x_j}(a_ib_j) = b_j\frac{\partial a_i}{\partial x_j} + a_i\frac{\partial b_j}{\partial x_j} \end{equation} If we now put $\vec{a} = \rho\vec{u}$, $\vec{b} = \vec{u}$ and assume that $\rho$ is a constant then we have $\nabla\times(\rho\vec{u}\vec{u}) = \rho\vec{u}\cdot\nabla\vec{u} + \rho\vec{u}\nabla\cdot\vec{u}$. If the fluid is incompressible, then $\nabla\cdot(\rho\vec{u}\vec{u}) = \rho\vec{u}\cdot\nabla\vec{u}$. Thus, \begin{equation} \iiint\nabla\cdot(\rho\vec{u}\vec{u})dV = \iiint \rho\vec{u}\cdot\nabla\vec{u}dV \end{equation} The integral on the left hand side can be transformed to surface integral and \begin{equation} \iint \rho\vec{u}\vec{u}\cdot\hat{n}dS = \iiint \rho\vec{u}\cdot\nabla\vec{u}dV. \end{equation}

Before proceeding, let us prove the identity $\nabla\cdot(\vec{a}\vec{b}) = \vec{b}\cdot\nabla\vec{a} + \vec{a}\nabla\cdot\vec{b}$. The derivation becomes easy when we use the cartesian tensor form, \begin{equation} \frac{\partial}{\partial x_j}(a_ib_j) = b_j\frac{\partial a_i}{\partial x_j} + a_i\frac{\partial b_j}{\partial x_j} \end{equation} If we now put $\vec{a} = \rho\vec{u}$, $\vec{b} = \vec{u}$ and assume that $\rho$ is a constant then we have $\nabla\cdot(\rho\vec{u}\vec{u}) = \rho\vec{u}\cdot\nabla\vec{u} + \rho\vec{u}\nabla\cdot\vec{u}$. If the fluid is incompressible, then $\nabla\cdot(\rho\vec{u}\vec{u}) = \rho\vec{u}\cdot\nabla\vec{u}$. Thus, \begin{equation} \iiint\nabla\cdot(\rho\vec{u}\vec{u})dV = \iiint \rho\vec{u}\cdot\nabla\vec{u}dV \end{equation} The integral on the left hand side can be transformed to surface integral and \begin{equation} \iint \rho\vec{u}\vec{u}\cdot\hat{n}dS = \iiint \rho\vec{u}\cdot\nabla\vec{u}dV. \end{equation}