How to find voltage across the 2k ohm resistor in the given circuit?
This is a linear circuit and the determination of \$I_x\$ is the first step. The \$2I_x\$ controlled source splits in \$I_x\$ in the 1-k\$\Omega\$ and the 2-k\$\Omega\$ resistors. The current is thus the voltage across the 2-k\$\Omega\$ resistor divided by \$R_2\$. If you do the maths ok, you should get \$I_x=\frac{I_xR_1-(I_xR_3-V_1)}{R_2}\$ with labels as shown in the below sheets. Solving for \$I_x\$ gives \$I_x=\frac{V_1}{R_2-R_1+R_3}=3\;mA\$. As \$I_x\$ circulates in \$R_3\$ also, the Thévenin voltage is simply \$V_{th}=R_3I_x=3\;V\$.

Now, short the 1-k\$\Omega\$ resistance and determine \$I_x\$ once again: \$I_x=\frac{R_1I_x+V_1}{R_2}\$. Solving for \$I_{sc}\$ gives \$I_{sc}=\frac{V_1}{R_2-R_1}=6\;mA\$. The Thévenin output resistance is thus \$R_{th}=\frac{V_{th}}{I_{sc}}=500\;\Omega\$.

A quick sim confirms the Thévenin voltage:


A .TF on a 1-A current source determines the small-signal output resistance \$R_{th}\$:



The .OUT file says:

***** SMALL SIGNAL DC TRANSFER FUNCTION  

output_impedance_at_V(2) 5.000000e+002
i1#Input_impedance       5.000000e+002
Transfer_function        5.000000e+002


Confirming the 500-\$\Omega\$ results.

The Mathcad file is below.