The standard transistor Schmidt trigger thresholds are set by combinations of resistors to be a fraction of the supply Voltage, but not necessarily .33 and .66

By choice of the resistor values both thresholds can be set and a variable resistor may be used to adjust them in use.

Because the rate of charge & discharge of the capacitor is also proportional to the supply Voltage, then the frequency of a Schmidt oscillator would be independant of the supply.

I don't know how CMOS implements the "Schmidt trigger" input, but, as Duke says, their trigger points are not necessarily proportional to the supply. (Though from my viewing of a few data sheets, they appear to be approximately so.)

WHY the frequency changes:

The time it takes for a capacitor to charge/discharge through a fixed resistor between any two fractions of the supply Voltage is the same. If a capacitor takes 1 msec to charge from 2V to 3V from a 5V supply, it will take the same 1 msec to charge from 4V to 6V on a 10V supply, because it needs twice the amount of charge to flow and the current is twice as much because the supply is twice as much. So the frequency of a relaxation osc will change with supply if the thresholds, as a fraction of the supply Voltage, change. If the thresholds are a constant fraction of the supply, the osc will be stable.

HOW the frequncy changes: depends on how the thresholds change.

If the % difference gets bigger, the charge time is longer, so the frequency is lower & vice versa. Say the thresholds were2V and 3V at 5V supply (20%), but 4V and 6.1V at 10V (21%), then the frequency would be lower at 10V.

There is a second factor to add, which is that the rate of charge decreases as the Voltage on the capacitor rises (and the rate of discharge decreases as the Voltage on the capacitor falls.) The net effect is that as the thresholds move away from the midpoint, the frequency falls. So 4.1V and 6.1V on the 10V supply would give a lower frequency than 2V and 3V on the 5V supply, even though the % difference is the same.

As for calculating these frequencies you'll have to consult Wikipedia or some such. I used to be able to do it, but my maths is getting rather rusty these days!

There are also other complicating factors. The simple calculation assumes we are charging to V+ and discharging to V-, but in a circuit like yours we are dis/charging from the output of the gate. For CMOS this output is pretty close to the supply rails and is symmetric, but for other technologies (TTL particularly) Vout Hi and Vout Low are not always symmetric.