How to use thevenin theorem in this example?
When you remove the load, you end up with:







simulate this circuit &ndash; Schematic created using CircuitLab

It follows that \$I_y=0\$ since it is an open circuit.

$$I_x=\dfrac{9\text{V}}{4\Omega+4\Omega+6\Omega}=0.643\text{A}$$

The potential difference between the terminals A and B is the \$V_{\text{TH}}\$. This is just the voltage across the farthest to the right \$4\Omega\$ resistor.

Then 

$$V_{oc}=V_{\text{TH}}=(0.643\text{A})(4\Omega)=2.57\text{V} $$

You need to find \$R_{\text{TH}}\$. Short circuit the 9V voltage source and find the equivalent resistance at the A and B terminals. That would be:

$$ R_{\text{TH}}=((4\Omega+6\Omega)||4\Omega)+5\Omega=7.86\Omega$$

With that, you can find \$I_2\$.