Why do you need to replace the inductor with a voltage source of 1V in this dependent source circuit to find its thevenin resistance?
What you are asked to do, provided I understand your request correctly, is to determine the resistance "seen" by the inductor when you temporarily disconnect it from its connecting terminals. A resistance or an impedance is a transfer function linking a response - the voltage \$V_T\$ across the current source - to a stimulus, the current source \$I_T\$. You will find more details here about these transfer functions. When you are asked to determine a resistance or an impedance from two terminals, install a test current generator \$I_T\$ and determine the voltage \$V_T\$ that appears across its terminals. The resistance is therefore \$R=\frac{V_T}{I_T}\$ or the impedance is \$Z(s)=\frac{V_T(s)}{I_T(s)}\$. The below sketch shows how to do it with your circuit:



Then, by invoking KCL and KVL, determine the relationship linking \$V_T\$ to \$I_T\$. If everything goes well, you should find \$R=\frac{R_1(R_2-3)}{R_1+R_2}\$. Please note that the 3 in this equation has the dimension of ohms. Applying your component values leads to a resistance of 0.333 \$\Omega\$ as confirmed by the dc operating point below. The test generator is 1 A so the voltage across its terminals divided by 1 A is the resistance you want in this linear circuit (Thanks Mr Morton!):



You would install a voltage source instead of a current source in case you would have to determine an admittance defined as \$Y=\frac{I_T}{V_T}\$. In this case, the response is the current \$I_T\$ while the stimulus is the voltage source \$V_T\$.