Assuming that the op amp and the diode are ideal (for op amp: infinite input impedance and 0 output impedance, for diode: 0 voltage drop):
D1 is open <=> $$V_i \frac{R2}{R1}>E$$
- case 1 (D1 closed):
$$V_i < E\frac{R1}{R2}$$
$$V_o = -V_i \frac{R2}{R1}$$ (It's a simple inverter)
- case 2 (D1 open):
$$V_i > E \frac{R1}{R2}$$
$$Vo = -V_i \frac{R2||R3}{R1} -E \frac{R2||R3}{R3}$$
How do you get the last relation (case2, Vo=...) ?
where $$R1||R2 = \frac{R1 \cdot R2}{R1+R2}$$
