You can't use Ohm's law because the diode is nonlinear. The idea is that the diode and the resistor share the same current, I, while their added voltage drops amount to 3V. This means, IR + V(I) = 3, where V(I) is the diode drop, a function of the intensity through it. Taking the Ohm's term to the other side of the equation, V(I) = 3 - IR. The right-side term is a straight line, with negative slope -R (note the scale in milliamps in the graphic, in that scale the slope is just -1, for R = 1KOhm.) That line intersects the y axis at I=3, V=0, and the x axis at V=3, I=0. When that straight line intersects the diode I-V characteristic, the equation V(I) = 3 - IR is satisfied, and you get the diode current and voltage at equilibrium. It's easy to see that V(I) = 1V, I = 2mA.
The circuit is quite like an inverting voltage adder, save for the diode. Using the load line analysis, we get a diode drop of 1V for a current of 2mA (I'd expect about 0.6V for a typical diode, though). So the current through the feedback resistor is 3 + 2 = 5mA, and the output voltage -5V.
If you try a spice simulation of the circuit, with a 1N4148 diode, you get an output voltage of -5.375V. The discrepance is due to a voltage drop of 0.625 V through the 1N4148, more typical than a 1V drop.