What you have is essentially this:
simulate this circuit– Schematic created using CircuitLab
All I've done is some re-drawing to finally get to the above-right.
The input voltage has a peak magnitude of \$30\:\text{V}\$ so the Thevenin of it has a peak magnitude of \$20\:\text{V}\$.
When \$-3\:\text{V}\le v_{\text{th}}\le +2\:\text{V}\$, then \$v_{\text{out}}=v_{\text{th}}\$ and just follows the input, as there is no current in either \$R_1\$ or \$R_2\$.
When \$v_{\text{th}}\le -3\:\text{V}\$, then \$v_{\text{out}}=-3\:\text{V}+\frac{1\:\text{k}\Omega}{1\:\text{k}\Omega+\frac23\:\text{k}\Omega}\cdot\left( v_{\text{th}}-\left(-3\:\text{V}\right)\right)=0.6\cdot v_{\text{th}}-1.2\:\text{V}\$. This \$v_{_\text{out}}\$ peak will be at \$-13.2\:\text{V}\$. There will be no current in \$R_1\$ since \$D_1\$ is off.
When \$v_{\text{th}}\ge +2\:\text{V}\$, then \$v_{\text{out}}=2\:\text{V}+\frac{1\:\text{k}\Omega}{1\:\text{k}\Omega+\frac23\:\text{k}\Omega}\cdot\left( v_{\text{th}}-2\:\text{V}\right)=0.6\cdot v_{\text{th}}+0.8\:\text{V}\$. This \$v_{_\text{out}}\$ peak will be at \$+12.8\:\text{V}\$. There will be no current in \$R_2\$ since \$D_2\$ is off.
(The above assumes ideal diodes with \$0\:\text{V}\$ voltage drop. You can work out the Thevenin resistance for each situation.)