$\nabla b \times \hat{z}$ using index notation $b$ is a scalar, $\hat z$ is the unit vector in the vertical direction and I need to evaluate a vector $$\vec{v} = \nabla b \times \hat{z}.$$
What I have done so far:
$$\vec{v} = \nabla b \times \hat{z} = \epsilon_{123}\frac{\partial b}{\partial x_2}e_3e_1 + \epsilon_{213}\frac{\partial b}{\partial x_1}e_3e_2 = \frac{\partial b}{\partial x_2}\delta_{31} - \frac{\partial b}{\partial x_1}\delta_{32} = \frac{\partial b}{\partial x_2} - \frac{\partial b}{\partial x_1}$$
which is definitely wrong because $\vec{v}$ should be a vector. It might be something obvious I can't find but I would like to know exactly each component of $\vec{v}$.

In Einstein summation its
$$ \vec v _i= \epsilon_{ijk} \partial_j b \hat z_k$$
so
$$ \vec v = \epsilon_{1jk} \partial_j b \hat z_k \hat x + \epsilon_{2jk} \partial_j b \hat z_k \hat y+ \epsilon_{3jk} \partial_j b \hat z_k \hat z$$
Since the only nonzero component of $\hat z$ is $\hat z_3 = 1$,
\begin{align}
 \vec v 
&= \epsilon_{1j3} \partial_j b \hat z_3 \hat x + \epsilon_{2j3} \partial_j b \hat z_3 \hat y+ \epsilon_{3j3} \partial_j b \hat z_3 \hat z \\
&=  \epsilon_{1j3} \partial_j b \hat x + \epsilon_{2j3} \partial_j b  \hat y+ \epsilon_{3j3} \partial_j b \hat z\end{align}
The $\hat z$ component is zero because $\epsilon_{3j3}= 0$ which we could have directly seen before any computation from the fact that $\vec a\times \vec b $ and $\vec b$ are perpendicular. Now since $\epsilon_{123}=1$ and $\epsilon_{213}=-1$ are the only relevant non-zero terms,
$$ \vec v = \partial_2 b \hat x  - \partial_1 b \hat y$$
It seems your mistake was saying that the third component of $\hat z$ is $e_3$, but a component of a vector is not another vector, it is a scalar.
