Thevenin Equivalent Network: How to treat Capacitor across 0v potential, with Dependent Sources?
Let's assume you have to find Thevenin's equivalent circuit and not Norton's. It'd be nice to find difference between them and I'm leaving that on you, however you need find Isc if you have to find Nortan's eq ckt and not in case you want Thevenin's eq one. 

In either case the procedure to find Rth (or Rn) and its value remains same. Now I have legit no idea why your professor wants you to treat capacitor as short circuit, it acts as one if frequency tends to infinity, but here everything looks DC to me and it has to act open.  Rth is resistance as seen from AB terminals, and when dependent sources are there, you simply apply some arbitrary source at AB and calculate how much current network is drawing and then ratio of applied source voltage to drawn current is your Rth. Of course, you have to set independent sources to zero first. Other method is to calculate quotient of Voc to Isc. 

Now to find Rth, apply Vab in such a way that your calculations will become easy, here lets say Vab = 25V, which is same as Vx hence Vx = 25V, this will give you 4x25 = 100V for dependent source. Remember to set independent sources to zero! Now, as capacitor acts as an open ckt, you have Ix = -(100/100) = -1A. Minus cause it's opposite to given direction.  Apply KCL at node A and you'll have 1.5A coming out from 25V source you connected and hence Rth is 25/1.5 Ohm. 

How will you find Voc now? Simple, capacitor acts as open, so you have two loops, two unknowns Vx and Ix with Voc= Vx and 2x2 matrix to solve! Cheers and welcome to ee.se!