Question on exterior differential $d\omega$ computation

The algorithm is: once you have $\omega$ written as a functional linear combination of the ${\rm d}x^I$'s, you apply ${\rm d}$ to the coefficients and replace the multiplication with wedge multiplication. In this case:

$$\begin{align*} {\rm d}\omega &= {\rm d}a_1 \wedge {\rm d}x \wedge {\rm d}y + {\rm d}a_2 \wedge {\rm d}x\wedge {\rm d}z + {\rm d}a_3 \wedge {\rm d}y\wedge {\rm d}z \\ &= \frac{\partial a_1}{\partial z} {\rm d}z \wedge {\rm d}x \wedge {\rm d}y + \frac{\partial a_2}{\partial y}{\rm d}y \wedge {\rm d}x\wedge {\rm d}z + \frac{\partial a_3}{\partial x}{\rm d}x \wedge {\rm d}y\wedge {\rm d}z \\ &= \left(\frac{\partial a_1}{\partial z} - \frac{\partial a_2}{\partial y} + \frac{\partial a_3}{\partial x} \right){\rm d}x\wedge{\rm d}y\wedge{\rm d}z,\end{align*}$$ using on the second line that $$\begin{align*}{\rm d}a_1\wedge {\rm d}x \wedge {\rm d}y &= \left(\frac{\partial a_1}{\partial x}{\rm d}x+\frac{\partial a_1}{\partial y}{\rm d}y+\frac{\partial a_1}{\partial z}{\rm d}z\right)\wedge {\rm d}x\wedge {\rm d}y \\ &= \frac{\partial a_1}{\partial x}{\color{red}{{\rm d}x}}\wedge {\color{red}{{\rm d}x}}\wedge {\rm d}y + \frac{\partial a_1}{\partial y}{\color{red}{{\rm d}y}}\wedge {\rm d}x\wedge {\color{red}{{\rm d}y}} + \frac{\partial a_1}{\partial z}{\rm d}z\wedge {\rm d}x\wedge {\rm d}y \\ &= \frac{\partial a_1}{\partial z}{\rm d}z\wedge {\rm d}x\wedge {\rm d}y, \end{align*}$$and similarly for the others (as wedge-repeats of $1$-forms kill).