Solution:

For two disjoint triangles to be formed, three of the balls must be connected into a triangle by three of the sticks, and the three remaining balls must be connected by the three remaining sticks.

There are $\binom{6}{3}$ ways to pick the 3 balls for the first triangle. Note that once we choose the 3 balls for the first triangle, the remaining 3 balls must form the vertices of the second triangle.

Now that we have determined the vertices of each triangle, we can assign the 6 sticks to the 6 total edges in the two triangles. Because any ordering of the 6 sticks works, there are $6! = 720$ total ways to assign the sticks as edges.

Finally, because the order of the two triangles doesn't matter (i.e. our initial choice of 3 balls could have been used for the second triangle), we must divide by 2 to correct for overcounting. Hence the final answer is $\binom{6}{3} \cdot 6! / 2 = 7200$.

Solution:

First, we ignore all the symmetries in the problem. There are then $6!$ ways to arrange the balls and $6!$ ways to arrange the sticks. However, each triangle can be rotated or reflected, so we have overcounted by a factor of $6^{2}$. Moreover, the triangles can be swapped, so we must also divide by 2. Thus the answer is
$$\frac{(6!)^{2}}{6^{2} \cdot 2} = \frac{120^{2}}{2} = 7200$$