Three. (20 points) As shown in Figure 2, $\triangle A O B$ is an equilateral triangle, and the coordinates of point $B$ are $(2,0)$. A line $l$ is drawn through point $C$ $(-2,0)$, intersecting $A O$ at $D$ and $A B$ at $E$, such that the areas of $\triangle A D E$ and $\triangle D C O$ are equal. Find the function expression of line $l$.
Three, $\because \triangle A D E$ and $\triangle D C O$ have equal areas,
$\therefore \angle A O B$ and $\triangle C B E$ have equal areas.
The area of $\triangle A O B$ is $\sqrt{3}$. Let the coordinates of point $E$ be $\left(x_{0}, y_{0}\right)$, then the area of $\triangle C B E$ is $2 y_{0}$. From $2 y_{0}=\sqrt{3}$, we get $y_{0}=\frac{\sqrt{3}}{2}$.

Also, the equation of line $A B$ is $y=-\sqrt{3}(x-2)$, and since $E$ is on $A B$, then $y_{0}=-\sqrt{3}\left(x_{0}-2\right)$. This gives $x_{0}=\frac{3}{2}$.
Thus, the coordinates of point $E$ are $\left(\frac{3}{2}, \frac{\sqrt{3}}{2}\right)$.
Since the coordinates of point $C$ are $(-2,0)$,
the equation of line $l$ is $y=\frac{\sqrt{3}}{7}(x+2)$.