I dont think this is your typical Thevenin problem you may have to reach a little to solve this. That is because you have to find a way to solve for Ix while also eventually eliminating Ix from the equation(s) so that you end up with just one voltage source and one resistance.
You should show your own work first, but because you didnt reply yet i might advise you to try solving this using the method of the loading of a linear network. That says that if you load a network with a resistance Rx and the voltage goes down to 1/2 of what it was when Rx is not connected, then Rx is the equivalent output resistance. You also have to solve for the output voltage while eliminating Ix.
This should be possible because it is a linear network, and a linear network will have a linear response.
If you get too stumped we can go over one such solution but it would be better if you posted your ideas about how to solve this first.
Maybe you got stuck, so another method is to just calculate the open circuit output voltage, then calculate the output current with the output shorted, then divide that open circuit voltage by that short circuit current and you have the Thevenin resistance.
You can find the output voltage by first finding the output voltage dependent on Ix, then since Ix is measured with R3 you divide that voltage by R3 to get a expression for Ix, then equate that to Ix. At that point you will have Ix on both sides of the equation, so then just solve for Ix explicitly. After that you can then plug the result for Ix back into the expression for the output voltage and that gives you the output voltage void of Ix. So now you would have the open circuit voltage
Next you would short the output and because that means that Ix=0 the output current should be easy to find. You then divide Vout by Iout and that gives you Rout which is Rth.