That was a dumb mistake in the Vo'' part by me. I think then it should be 2V, since there's no current for the dependant source to be on. So the overall Voc would be 6V, is that correct?Actually, I owe you an apology. I thought you were doing something different than you actually were. What you did in order to find Rth is just fine (except you should learn to track your units properly). Rth=4Ω.
Having recognized that, I then looked at your second image and your work for finding Vth is close, but not quite there. Take a look at Voc'' more closely.
My comments regarding whether the current of Ix=1A apply to a 1A source applied to the terminals of the entire circuit (i.e., with both independent sources on). In that case, you have a 1A current source driving current into a 4Ω resistor that is in series with a 6V supply (i.e., the Thevenin equivalent circuit -- so there's a free hint) resulting in a terminal voltage of 6V+1A*4Ω=10V.
You then have Ix=2A which makes the voltage at negative terminal of the dependent source 10V-(2V/A)*Ix=6V. You then drop another 2A*2Ω=4V going across the 2Ω resistor, leaving you with 2V at the top of the 2V source, so everything works out.
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