The base BC of an equilateral triangle ABC lies on y-axis. The coordinates of point C are (0, −3).

Let (0, y) be the coordinates of B. Then

$0=\frac{-3+y}{2} \Rightarrow y=3$

Thus, the coordinates of B are (0, 3).
Here, AB = BC = AC and by symmetry the coordinates of A lies on x-axis. Let the coordinates of A be (x, 0). Then

$A B=B C \Rightarrow A B^{2}=B C^{2}$

$\Rightarrow(x-0)^{2}+(0-3)^{2}=6^{2}$

$\Rightarrow x^{2}=36-9=27$

$\Rightarrow x=\pm 3 \sqrt{3}$

If the coordinates of point $A$ are $(3 \sqrt{3}, 0)$, then the coordinates of $D$ are $(-3 \sqrt{3}, 0)$.

If the coordinates of point $A$ are $(-3 \sqrt{3}, 0)$, then the coordinates of $D$ are $(3 \sqrt{3}, 0)$.

Hence, the required coordinates are $A(3 \sqrt{3}, 0), B(0,3)$ and $D(-3 \sqrt{3}, 0)$ or $A(-3 \sqrt{3}, 0), B(0,3)$ and $D(3 \sqrt{3}, 0)$.