We only talk about exterior derivatives of differential $k$-forms, not vector fields. However, what we can do is the following: given a vector field $F: \Bbb{R}^3 \to \Bbb{R}^3$, $F = (F_x, F_y, F_z)$, we can consider the following one-form:
\begin{align}
\omega &= F_x \, dx + F_y \, dy + F_z \, dz
\end{align}
And yes, the exterior derivative of the one-form $\omega$ is indeed the thing you wrote down:
\begin{align}
d\omega &= \left(\dfrac{\partial F_y}{\partial x} - \dfrac{\partial F_x}{\partial y}\right) dx \wedge dy + \left(\frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x} \right) dz \wedge dx + \left(\frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z} \right) dy \wedge dz
\end{align}
Just some fun extra tidbits: if you know some vector calculus, the above expression probably looks pretty familiar, almost like the curl of $F$, though not quite.
If you want to somehow get the curl of $F$ from here, you need to look at the "Hodge star" operator, which assigns to the above $2$-form $d\omega$ a certain $1$-form $\alpha$, namely
\begin{align}
\alpha &= \left(\dfrac{\partial F_y}{\partial x} - \dfrac{\partial F_x}{\partial y}\right) dz + \left(\frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x} \right) dy + \left(\frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z} \right) dx
\end{align}
then from here, you can get a vector field, $G$, (pretty much by replacing $dx$ with $e_x$, $dy$ with $e_y$ and $dz$ with $e_z$),
\begin{align}
G:= \left( \frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z}\right) e_x + \left(\frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x} \right) e_y + \left(\dfrac{\partial F_y}{\partial x} - \dfrac{\partial F_x}{\partial y}\right) e_z,
\end{align}
and this is precisely the curl of $F$
