Combination circuit - how to calculate equivalent capacitance when switch is closed?
Yes, @michal-podmanicky is right, from a static point of view the result is 4.5uF.
When you short circuit the switch (status = closed, switch ideal) you have two parallel cap = 6+3 uF, and then each group of 2 is in series : 9//9 = 9/2 = 4.5uF.
Maybe you wanted to know what happens during the transient of closing the switch? I ask because you speak of potential difference and not equivalent capacitance. I this case you need to provide the external circuit (e.g. voltage source and equivalent resistance) and the switch resistance. It can be solved as a Wheatstone bridge, where capacitors are in place of the resistors you usually see.