Directional derivative of a vector field

Just to fix some notations, let $$X\colon\mathbb{R}^3\to\mathbb{R}^3,\quad X(x_1,x_2,x_3) = \begin{pmatrix}X_1(x_1,x_2,x_3)\\ X_2(x_1,x_2,x_3)\\ X_3(x_1,x_2,x_3)\end{pmatrix}.$$ Define the Jacobian of $X$ as $$ J\in\mathbb{R}^{3\times 3},\quad J_{ij}=\frac{\partial X_i}{\partial x_j}=\partial_jX_i,\quad i,j=1,2,3. $$ and the third order tensor $$ H_{kij} = \frac{\partial^2 X_k}{\partial x_i\partial x_j} = \partial_i\partial_jX_k,\quad i,j,k=1,2,3. $$ This means that, for each $u,v\in\mathbb{R}^3$, $$ (H[u,v])_k = \sum_{i=1}^3\sum_{j=1}^3 \frac{\partial^2 X_k}{\partial x_i\partial x_j}u_iv_j $$

Now we can rewrite $Y=(X\cdot\nabla)X$ as $$ Y = JX,\quad\text{i.e.}\quad Y_i= (JX)_i = \sum_{j=1}^3\frac{\partial X_i}{\partial x_j}X_j,\quad i=1,2,3. $$ We want to compute $(Y\cdot\nabla)Y$. First we must compute the jacobian of $Y$. For any $i=1,2,3$, $$ \partial_jY_i = \sum_{k=1}^3\frac{\partial^2X_i}{\partial x_j\partial x_k}X_k+\sum_{k=1}^3 \frac{\partial X_i}{\partial x_k}\frac{\partial X_k}{\partial x_j} $$ The second term is given by the matrix product $(JJ)_{ij}=(J^2)_{ij}$. The first term is actually given by $$ \sum_{k=1}^3\frac{\partial^2X_i}{\partial x_j\partial x_k}X_k=\sum_{k=1}^3X_k\,\partial_j\partial_k X_i = (H[e_j,X])_i $$ where $e_j$ is the versor in the direction $j=1,2,3$. This means that we can write $$ \partial_jY_i=(H[e_j,X])_i+(J^2)_{ij} $$ From this, it follows that $$ (Y\cdot\nabla)Y=(\nabla Y)Y=H[JX,X]+J^2(JX). $$ We observe that the first term is indeed the correct one because $$ (Y\cdot\nabla)Y_k=Y_k\sum_{i=1}^3\sum_{j=1}^3 (X_j\,\partial_i\partial_j X_k + \partial_iX_j\,\partial_jX_k) $$ and we see that $$ \sum_{i=1}^3\sum_{j=1}^3 Y_k\,X_j\,\partial_i\partial_j X_k=(H[Y,X])_k=(H[JX,X])_k $$