LM385Z-2.5 in series with voltage divider - voltage reference problem
The load current across the resistors parallel to the references, assuming you have 5 volts, is \$\frac {5V}{6.9k\Omega}  =0.724mA \$.  When you put that current through your \$10k\Omega\$ resistor, the voltage drop across that is 7.24 Volts.  Add that to your 5Volts, and you have 12.24V, which you're trying to provide with a 9V battery.  This, of course, ignores your bias current through your LM385-2.5's.

You can't do this with one 9V battery.  You need to drop the 10K to a lower value, or  use two batteries, but make sure you don't exceed the maximum bias currents for your references.