I am still not understanding how to calculate n in Ecell=Enaughtcell-(.0592/n)logQ.

Question

Example: Au/Au^3+ and Cu/Cu^2+
If I am correct it should look like this
Au->Au^3+ + 3e- (Enaughtcell = 1.50V) and Cu->Cu^2+ + 2e- (Enaughtcell = 0.34V). The Cu/Cu2+ is the less negative value so that rxn gets reversed. The Au 1/2rxn is losing e- so its oxidized and the Cu 1/2rxn is gaining e- so its reduced. I multiply the Au rxn by 2 and the Cu rxn by 3 to balance the charges. Is this correct so far? If this is the case why is n not equal to 6?

You may be confusing yourself with the Enaughtcell thing. The values you are quoting are half cells and not cells. I think you would do better to simply call these Enaught (for E^o) and not put cell on it until you have added the half cells together to get the cell.
If I look up the values in my tables I don't get the same values you have. I find:
Au³⁺ + 3e ==> Au(s) E^o = +1.5 volts

Cu²⁺ + 2e ==> Cu(s) E^o = +0.34.

The less negative value gets reversed so
Cu(s) ==> Cu²⁺ + 2e E^o_oxdn = -0.34
Au³⁺ + 3e ==> Au(s)
E^o_redn = +1.5 volts.

Now do the 3x and 2x and add the equations.
E^o cell = E^o_oxdn+ E^o_redn.
The rxn then is
3Cu + 2Au³⁺ ==> 3Cu²⁺ + 2Au

The Cu is losing electrons and the Au is gaining electrons.

As for n, the change in electrons, for the complete reaction it is 6 and you can do that one of two ways.

Procedure 1.
Use E = E^o - 0.0592/n log Q for each half reaction. n will be 2 for Cu and 3 for Au. then add the two E values to obtain E cell (not E^o cell).

Procedure 2.
Add E^o values as I did above, obtain E^o for the cell, THEN use E cell = E^o - 0.0592/n log Q to obtain E cell. In this case n = 6 as you have in your example.

Check all of this very carefully and check my thinking. It is so easy to make an error in typing with all these subscripts and superscripts. If I have made an error I will try to get back and post a correction. It may be better not to use all the subscripts and superscripts. We'll see.