Notice that throughout this question I will be using Einstein's summation notation.
Let $v(x):\mathbb{R}^3 \to \mathbb{R}^3$ be a vector field and define
$$2\mathbb{W}(v) = \nabla v - \nabla v^T$$
where
$$\nabla v = \frac{\partial v_i}{\partial x_j}e_i \otimes e_j$$
and where $$a \otimes b$$ is the matrix where the entry $(i,j)$ is $a_i b_j$.
I am asked to show
$$2\mathbb{W}(v)(x - x_0) = (\nabla \times v) \times (x-x_0)$$
If I do it component-wise, I am able to do it with no problem. Because for this course we have been using Einstein's summation notation a lot, I decided to try and do it myself. For the LHS I got to
$$2\mathbb{W}(v)(x - x_0) = \left(\frac{\partial v_i}{\partial x_j} - \frac{\partial v_j}{\partial x_i} \right)(x-x_0)_je_i$$
which agrees with my result when I did everything component by component.
[Write $d = (x-x_0)$ for the sake of brevity]. For the triple product, I used the relation $(\nabla \times v)\times d = (d\cdot \nabla)v - \nabla (d\cdot v)$ where I got
$$(d\cdot \nabla)v = d_j\frac{\partial v_i}{\partial x_j}e_i $$
and
$$\nabla(d\cdot v) = \frac{\partial}{\partial x_i}(d_jv_j)e_i = \left(\frac{\partial d_j}{\partial x_i}v_j + d_j\frac{\partial v_j}{\partial x_i} \right)e_i$$
which, when put together, yields
$$\left(\frac{\partial v_i}{\partial x_j} - \frac{\partial v_j}{\partial x_i}\right)d_je_i - \frac{\partial d_j}{\partial x_i}v_j e_i$$
which has that extra $\frac{\partial d_j}{\partial x_i}v_j e_i$... I have no idea where my mistake is! Thank you for your help.