No Kane, you do not appear to be understanding the circuit.
When the supply voltage is below the Zener voltage, the Zener is off and the capacitor is charged to whatever the voltage is. If a load is applied, the voltage will decrease in proportion.
When the supply voltage is above the Zener voltage, the Zener is on and current flows through it. The capacitor is charged to the Zener voltage.
If a load is applied, current flows through the load and the Zener current decreases. But the voltage does not decrease very much. So the voltage is regulated at the Zener voltage (about 5.1 Volt)
Provided that the current taken by the load is less than the current through the Zener before the load was applied, the voltage will be regulated. Once the load current exceeds this value, the voltage will decrease in proportion.
Your circuit shows that the supply voltage is 9 Volt and the resistor is 10k. Thus with no load, the Zener current will be about 9 - 5.1)/10k = 3.9/10 = 3.9 mA.
So the voltage will be regulated at about 5.1 Volt provided the load is less than 3 mA. At a load of 3 mA there will be 0.9 mA through the Zener thus it is still on. But if the load increases, the voltage will decrease gradually until the load current reaches 3.9 Volt and then it will decrease in proportion thereafter.
The capacitor is provided in order to provide a low AC source impedance. It does not provide regulation.
Len