This looks like a linear low dropout regulator (LDO). I'm assuming that all transistors are PMOS except the Vx input transistor that is NMOS and all in saturation. Vtune is a current source that will change the amount of voltage shift that Vout will have. The NMOS with input x is operating as a common source amplifier, then the PMOS will add a level shift of Vgs2 proportional to Vtune. Changing the current using Vtune will change the Vgs2, so the final Vout=Vx+Vtn1+Vgs2. The idea is to make the (Vgs-Vtn) of the big transistor as small as possible so you have a low dropout between its drain and source.
The outer circuit is just to show the loop used. Assuming the big transistor is a common drain NMOS , you will have an inversion in the loop due to the common source NMOS and since the level shifter has a gain ~1, you have a negative feedback loop. The idea is to make Vtune=Vx while adding high current capabilities at the load connected at Vx. The level shifter is needed since you want to make Vds=(Vout-Vx-Vtn), then adding the Vgs2, you make sure that the transistor is always operating in saturation for a reasonable amount of current and with a small Vds.