7. As shown in Figure 2, points $A$ and $C$ are both on the graph of the function $y=\frac{3 \sqrt{3}}{x}(x$ $>0)$, points $B$ and $D$ are both on the $x$-axis, and such that $\triangle O A B$ and $\triangle B C D$ are both equilateral triangles. Then the coordinates of point $D$ are $\qquad$.
7. $(2 \sqrt{6}, 0)$.

As shown in Figure 9, draw perpendiculars from points $A$ and $C$ to the $x$-axis, with the feet of the perpendiculars being $E$ and $F$ respectively. Let $O E=a, B F=b$, then $A E = \sqrt{3} a, C F = \sqrt{3} b$.
Thus, $A(a, \sqrt{3} a)$,
$$
C(2 a+b, \sqrt{3} b) \text{. }
$$

Therefore, $\left\{\begin{array}{l}\sqrt{3} a^{2}=3 \sqrt{3}, \\ \sqrt{3} b(2 a+b)=3 \sqrt{3} .\end{array}\right.$ Solving these equations yields $\left\{\begin{array}{l}a=\sqrt{3}, \\ b=\sqrt{6}-\sqrt{3} \text{. }\end{array}\right.$ Hence, $D(2 \sqrt{6}, 0)$.