How can i switch 9VDC / 3VDC
There's 2.4 kA = 2400 A flowing which is more than &quot;leaking&quot;.
This current is flowing because there's a diode inside each MOSFET between Drain and Source (Actually these diodes are between Drain and Bulk but in discrete MOSFETs Bulk and Source are shorted). Here the diode in the MOSFET on the right (3V side) is in forward mode.
I've re-drawn your circuit and added the internal MOSFET Drain-Source diodes:



simulate this circuit &ndash; Schematic created using CircuitLab
Here M1 is switched on so it conducts and it tries to pull the its Drain to 9 V. This fails as there is a diode present in M2 that starts to conduct.
Even then the 2.4 kA is ridiculous but if the models of the MOSFET is simple and allows for that then this can happen. I never use Falstad circuit simulator as I have access to more advanced simulators (and so do you, try LTSpice or QUCS).
What you will need to do to solve this is to add a 2nd MOSFET in (anti) series with the MOSFET on the right (3V side), like this:



simulate this circuit
Note how the &quot;new&quot; MOSFET M3 is connected in &quot;anti series&quot; meaning that the Drains of M2 and M3 are connected together. This combination behaves as a MOSFET without the Drain-Source diode.
Now when the 9 V is enabled both M2 and M3 will be off and M3 will prevent any current flowing through the diode inside M2.
At the 9 V side a &quot;double MOSFET&quot; isn't needed as this MOSFET is connected to 9 V  which is the highest voltage in the circuit so there is no way that the diode D1 can start to conduct (you'd need more than 9 V at the Drain of M1).