I'm bothered if my circuit is correct. I've attached my circuit.

If a 9V external source is present, it will be the one powering the system while charging my battery. However, if the 9V is removed the system will be automatically supplied by the battery. Is my circuit correct? Does it have an effect on the MOSFET if current flows for drain to source everytime the battery powers the system?

 

You don't need the MOSFET.

Replace the MOSFET with a Schottky diode, to prevent the 9v source from charging your batteries. After that, you won't need R1 either.

 

Although Sarge is basically correct, there are other possible issues.

1) There is no ground symbol at the bottom of BT2, without a return the batteries/charger won't be powering anything.

2) The batteries will supply 7.2V at best -- make sure the difference between the battery voltage and 9V is not a problem for any devices being powered.

3) The output voltage will be ~8.5V or ~6.8V, but voltage drop across the diode will vary with the diode and current draw (be sure the diodes are rated 2x the max current that can be supplied).

4) Depending on the charger design -- the batteries may back-power the charger, which could drain the batteries quicker. Should this be the case, add a blocking diode (similar Schottky) between the charger and BT1.

Good Luck!

 

Simulate the whole thing with a good SPICE simulator such as the free SwitcherCad III from Linear Tech. SPICE simulation for analog electronics highly rewards those that learn it as you would see by yourself where the circuit fails.