Find the Thévenin equivalent circuit with the output terminals A and B
To find \$R_{th}\$, connect a fictitious voltage source of 1V between A and B, so that \$V_x = 1\$




Now find I using KVL, 
$$-2 + 6 - 1- 2I = 0$$
which means, \$ I = 1.5 A \$

Therefore, the thevenins resistance \$R_{th}\$ seen from AB will be just \$ V_x/I = 1/1.5 = 0.667 ohms \$


To find \$ V_{th}\$, look at the original ckt. It is the open circuit voltage across AB = \$V_x\$




\$V_x\$  should be equal to  \$-2V_x + 6\$  because current through 2 ohms = 0.

$$ V_x = -2V_x + 6 $$
$$\implies V_x = 2V  = V_{th} $$


Now you can draw your thev.eq.ckt using \$ R_{th} = 0.667A \$ and \$ V_{th} = 2V \$