I'm unsure what you mean by "VTH" in relation to the circuit pictured. However you can simplify the circuit to help you, have a think about how many resistors there really need to be. That will make any maths much easier.
Basically you are replacing voltage sources with shorts, and current sources with opens.
Start by opening (removing) the current source. The circuit decomposes into 2 simple resistive divider loops, one at the top of the voltage source, one at the bottom. Calculate the voltage drops around the circuits per Kirchoff and Ohm. Write them down.
Next, replace that voltage source with a short circuit. Now calculate the equivalent resistance as seen from the current source, make it a 2-resistor divider so the voltage source will be in the middle. Now calculate the voltage drops on that (hint: the top one will be 12V). No go back and add these drops algebraically to the voltages given in the previous step. The two resistors on the right (top and bottom) are in series, so they will have to be treated as dividers and re-calculated. If you have done this correctly, the three voltage drops of the resistors around the top and bottom of the 115V source will add to 115V.
Strictly speaking, you need a pair of output nodes to calculate the Thevenin voltage and resistance equivalents. You have not specified those.