Vth
Following along with what you already wrote, here's your schematic (on top) with part identifications added:

simulate this circuit – Schematic created using CircuitLab
By using the Norton to Thevenin conversions for both current sources, it quickly becomes the what you see on the bottom case above.
This is easy to solve using nodal analysis for \$V_A\$ now:
$$\begin{align*}\frac{V_A}{R_{\text{th}_1}+R_2}+\frac{V_A}{R_{\text{th}_2}}&=\frac{V_A-3\:\text{V}}{R_{\text{th}_1}+R_2}+\frac{1\:\text{V}}{R_{\text{th}_2}}\\\\\therefore\\\\V_A&=\frac{R_{\text{th}_1}+R_2-3\cdot R_{\text{th}_2}}{R_{\text{th}_1}+R_2}=-500\:\text{mV}\end{align*}$$
Therefore, \$V_\text{TH}=-500\:\text{mV}\$.
Rth
We can re-write this schematic in the following way, applying \$I_x=V_A-1\:\text{V}\$ so that we now know: \$I_x-2\:\text{V}=V_A-3\:\text{V}\$:

simulate this circuit
Since the voltage difference across \$R_{\text{th}_1}+R_2\$ is fixed (it cannot change and will always have the same magnitude), we can go further:

simulate this circuit
Node \$X\$ doesn't matter, as it is behind a current source which has \$\infty\:\Omega\$. Because a current source has infinite impedance, it doesn't affect the Thevenin impedance of this result. So this leaves just \$R_{\text{th}_2}\$ as representing the final impedance.
Therefore, \$R_\text{TH}=1\:\Omega\$.
Summary
So here is the result:

simulate this circuit