Suppose we have two scalar valued functions $a,b : \mathbb{R}^3 \to \mathbb{R}$ and wish to evaluate $\nabla (\nabla a \cdot \nabla b)$. We can begin with the formula
$$\nabla(A\cdot B) = (A\cdot\nabla) B + (B\cdot\nabla) A + A\times(\nabla \times B) + B \times (\nabla \times A)$$
for vector fields $A,B : \mathbb R ^3 \to \mathbb R ^ 3$, substituting $A=\nabla a$ and $B = \nabla b$ to find
$$\nabla (\nabla a \cdot \nabla b) = (\nabla a \cdot \nabla) \nabla b + (\nabla b \cdot \nabla) \nabla a$$
(I am confident that the last two terms will cancel because they involve the curl of a gradient. If I wanted to justify it rigorously, would I have to say that $d^2 = 0$? Or, would $\nabla \times \nabla = 0$ suffice?) To my understanding the wikipedia article from which I pulled the first identity defines
$$(A\cdot \nabla) = A_1 \partial_x + A_2 \partial_y + A_3 \partial_z$$
where $A_1, \dots, A_3$ denote the components of $A$ and $\partial_x, \dots, \partial_z$ are the partial derivatives in each of the three spatial coordinates.
I believe that in the first formula, the gradient of the vector fields $A$ and $B$ is defined by
$$ \nabla\vec{\bf F}=\left( {\begin{array}{c} \frac{\partial F_1}{\partial x}&\frac{\partial F_1}{\partial y}&\frac{\partial F_1}{\partial z}\\ \frac{\partial F_2}{\partial x}&\frac{\partial F_2}{\partial y}&\frac{\partial F_2}{\partial z}\\ \frac{\partial F_3}{\partial x}&\frac{\partial F_3}{\partial y}&\frac{\partial F_3}{\partial z}\\ \end{array}} \right) $$
but I am not sure if this is correct. (I might have the transpose of the correct answer.) $(\nabla a \cdot \nabla)$ and $(\nabla b \cdot \nabla)$ should act component wise, so I believe that
$$ (\nabla a \cdot \nabla) \nabla b = \begin{pmatrix} a_x b_{xx} + a_y b_{xy} + a_z b_{xz} \\ a_x b_{yx} + a_y b_{yz} + a_z b_{yz} \\ a_x b_{zx} + a_y b_{zy} + a_z b_{zz} \\ \end{pmatrix}= \begin{pmatrix} b_{xx} &b_{xy} & b_{xz} \\ b_{yx} &b_{yz} & b_{yz} \\ b_{zx} &b_{zy} & b_{zz} \\ \end{pmatrix} \begin{pmatrix} a_x \\ a_y\\ a_z \end{pmatrix} = (\nabla^2 b) \nabla a $$
where $\nabla^2$ sadly does not denote the Laplacian. (If the transpose is used, I believe then that $(\nabla a \cdot \nabla) \nabla b = \nabla a (\nabla^2 b)$.)
Question: Is my work correct? Is there any simpler expression that doesn't involve tensors? Could I try to argue in terms of differential forms and use the Hodge Laplacian? If I used $\Delta = (d + \delta)^2$ how would I express the original problem in the language of differential forms? Would that even assist me, or would the tensors just be hidden from my direct language?