My problem with that is that the other side of the diode on the Arduino is normally powered up. So for example if you have a 12V supply in the power jack you'll have 11.3V at Vin (because of the diode drop). Thus you are pumping 11.3V into your 9V battery which won't be designed to handle it. It will probably get hot and explode.
You would need another diode specifically between the battery and Vin. And make sure that Vin is at least 10V so that the normal mains power does not charge the battery.
Bear in mind a 9V battery doesn't have much capacity. With the power LED on, taking sensor readings, and sending data via Ethernet (if that is what you are doing) you may have only bought yourself 15 minutes.
I have a post about a temperature and humidity sensor which runs all the time from 3 x AA batteries. I think the batteries need replacing about every two years. The data is written to an SD card which you remove and plug into your computer when you want to.
when you have 9V hooked up to Vin and 5V coming in from the USB, it will use power from the 9V battery
Looks like I misread the question. @mwwalk is quite right. If you supply more than 6.6V through Vin then the circuitry switches the USB off. See here:
Vin is provided by the power jack (after a diode which drops it by 0.7V) - see blue ellipse. In any case (power jack or not) Vin goes through a voltage divider (two 10k resistors to the left of the red circle). This is compared to the output of the 3.3V voltage regulator. Therefore
GATE_CMD is HIGH if Vin exceeds 6.6V. This turns off the MOSFET (green circle) and disconnects the USB power.