So far you have solved the circuit twice. Once with is2 replaced with a short to calculate isc, and again with Vs replaced with a short to calculate Rth. You could now solve for the voltage across is2 in the original circuit using the network solving techniques you know. Or, you could do it using the Thevenin equivalent you obtained. Look at the last circuit in your working. You have a -7.31 volt source in series with a -161 Ω resistor and then in series with a current source of is2 amps. The current source will force a current of is2 amps through the -161 Ω resistor, so what will be the voltage across that resistor? That will make the value of V1 equal to the sum of -7.31 volts and the drop across the -161 Ω resistor. That gives you an expression for V1 in terms of is2.