While everything shown is correct, you need to take in mind what type of regulators they are. The 5v is a common LM7805 1.5A regulator, and the 3.3v is a LD1117V33 800mA low-drop out regulator, similar to the LM78xx family. Both are Linear Regulators.
First, the LM7805 is not a low-drop out regulator. It needs at least 2v above it's output to work properly. At 7.2v Battery input, that will quickly drop, depending on the current draw and battery discharge rate. A low dropout version would be better. And then I noticed the Diode. Any common diode will have 0.6 to 0.8v drop across it, as current increases beyond anything significant (10~100ma). This would make the Battery 7.2v way below the LM7805's Minimum Input Voltage, and it will not be happy. A lower dropout version would be mandatory for proper operation in this case.
Second, as linear regulators, they will turn any voltage above it's output, into heat. If you power them from 9V, and draw say 250mA each, that is (9V - 5V) * 250mA = 1 Watt of power wasted into heat for the LM7805, and (9V - 3.3V) * 250mA = 1.43 Watts on the LD1117V33. Heat will start becoming a concern, especially for the 3.3v regulator. A better option would be to have the 3.3v Regulator's input coming from the 5v regulator's output. While this means the 5v Regulator's load will increase by whatever load the 3.3V regulator consumes, the overall affect is that the 3.3V regulator will produce less heat.
Be sure to use a heatsink on the 7805, and watch your current draw.