Capacitor's energy required for a predetermined drop in its voltage - with LDO and constant current elements as load
I'm now sure if I understand your question right. I think you want to know (based on the capacitor voltage drop) how much energy is absorbed by the "always on" (static) circuit and how much by the MCU (dynamic) circuit? 
If you only know the voltage difference of the capacitor (dV) from t=0ms to t=10ms, then this is not possible to calculate, because you have two degrees of freedom in the equations. 
You would need to measure the capacitor voltage drops of the dynamic or the static circuits each alone.  

Also, like Andy aka wrote, your formula is not right: The right one would be 
$$dE=\frac{1}{2}CV_{0ms}^2-\frac{1}{2}CV_{10ms}^2$$