Solving Thevenin and Norton Circuit
To determine the Thévenin voltage \$V_{th}\$, express the voltage \$V_x\$. It can be obtained by writing: \$V_x=-R_x(2V_x-I_1)\$. In this expression, 2 is a conductance and its unit is the siemens. Solving for \$V_x\$ gives \$V_x=\frac{I_1R_x}{1+2R_X}=333\;mV\$. We can now immediately determine the Thévenin voltage across \$R_L\$ which is considered infinite here: \$V_{RL}=2V_xR_1+V_1=6.33\;V\$. A quick sim shows this is the correct answer:


To obtain the output resistance \$R_{th}\$ also called the incremental resistance or Thévenin resistance, we turn all excitation sources off and we "look" through the output terminals (across \$R_L\$ connecting terminals) to determine the resistance we see. Exactly as when determining time constants with the fast analytical circuits techniques (FACTs):



By inspection, we see that the resistance is \$R_1\$ and equal to \$2\;\Omega\$. To verify our answer, we can ask SPICE to calculate the transfer function using the first schematic diagram linking V(4) to source \$V_1\$ and invoking a .TF primitive function:

***** SMALL SIGNAL DC TRANSFER FUNCTION  

output_impedance_at_V(4)2.000000e+000
v1#Input_impedance      1.000000e+020
Transfer_function       1.000000e+000


We could also determine the short circuit current \$I_{sc}\$ and a quick sim tells us this current is 3.166 A:



The resistance we want is obtained by dividing \$V_{th}\$ by the short circuit current and we find our \$2\;\Omega\$ again.