Since we are finding the Norton with respect to the 3 Kohm, we take the 3 Kohm out of the circuit and consider the resistance seen from the terminals where the 3K was.
The circuit to the left of the 3K is already a Norton equivalent, where the Norton current is -10 mA (because it is facing down). The resistance is infinite. That is, when you open the current source to deactivate it, the 1K and 2K are left disconnected.
The circuit to the right of the 3K is already a Thevenin, where the voltage is 6V and the equivalent resistance is 9 Kohms. Converting to a Norton, we get Norton current of 0.667 mA and a resistance of 9 Kohms.
Now combine the two Nortons. The total curent will be -10mA + 0.667mA = -9.33 mA. The total resistance is infinite in parallel with 9K, which is simply 9K.
© 2010, Steven H. VanderLeest