Thevenin and Norton Equivalents
Your theory is sound, just a few minor errors.

For your 1st approach, to compute the norton current, you short the two terminals of the circuit you wish transform (A &amp; B for your circuit). The short circuit current is the same current through the output 5 V voltage source. When the output is shorted, the intermediate node you labeled as Vy in later analysis must be 5 V. Apply KCL at you will find 
$$ I_N = -4 \textrm{ mA} $$.

For the 2nd approach, your KCL at node \$V_y\$ is correct just an aritmetic error on the solution. $$ V_y = 3\textrm{ V} $$ and the output thevin voltage is then $$ V_{th} = -2 \textrm{ V} $$

For the third approach the two 1k resistors in parallel form an equivalent resistance of 500 ohms not 2 kOhms, which yields the same thevin voltage as
$$ V_{th} = -2 \textrm{ V} $$

The complete Norton/Thevin solution is then,







simulate this circuit &ndash; Schematic created using CircuitLab