Inverting boost converter by using MC33063A/MC34063A: Where does the IC get ground connection?
Firstly you have to recognize the basic topology of an inverting buck-boost circuit: -

There is a block labelled PWM (red) and that drives the switching element. In the MC33063A it is a bipolar transistor. During the first half of the switching cycle (BJT activated), current ramps up in the inductor L. That current is flowing through the inductor to ground. In the second half of the switching cycle that current has to continue to flow but is now diminishing in magnitude. The inductor is tied to ground at its lower point therefore its upper point has to become negative to ensure that current flows albeit diminishing.
The only path that this can happen through is via the diode hence, the output voltage (\$V_{OUT}\$) has to become negative.
The MC33063A's first role is to &quot;find&quot; a suitable power supply rail AND then deliver PWM to the switching element. So, it finds a suitable rail by connecting between \$V_{IN}\$ and \$V_{OUT}\$. Initially (at the start) \$V_{OUT}\$ is close to 0 volts but that isn't going to prevent the MC33063A from starting-up. As \$V_{OUT}\$ acquires the required negative voltage, the chip is still powered so there's no problem there; it can still deliver a PWM signal to the switch perfectly well with any reasonable value for \$V_{OUT}\$.
I suppose the subtler bit might be the 1.25 volt reference - one end sits at \$V_{OUT}\$ and therefore the comparator in the chip sits at \$V_{OUT}+\text{1.25 volts}\$. And so the comparator resistors R1 and R2 have to be chosen so that 1.25 volts is dropped across R1. That will be the regulation point for the chip.
For the -12 volt circuit in the question, the voltage magnitude across R1 is this: -
$$\text{12 volts}\cdot\dfrac{R_1}{R_1 + R_2} = \text{1.2494 volts}$$

My question is, how does the IC connect itself to the ground?

It uses the negative rail as a &quot;ground&quot; point for the chip.