The one in red circuit. How is it derived ?

 

This is a two-stage opamp. The first stage is differential pair with active load current mirror and the second stage is common source amplifier with current source load.
The total gain will be the product of gain of each stage. To find the gain of each stage, you could draw signal model or do it by inspection.

 

Before you start to study this type of a circuits, you already should know and and understand the gain expression for an differential amplifier and common source amplifier.
Because the gain expression in you picture is not the exact solutions of a full small-signal model. The formula they show on the picture is just a simplified version. They assume that the differential amplifier gain is -gm*RL (for the differential output or current mirror as a load ) and for common source amplifier they to the same think -gm*RL. And this is why we can do it by inspection.
The load resistance for a differential amplifier is ro2 and ro4 and the gain is - gm*(ro2||ro4). For CS amplifier (M5) the load is ro5 and ro6 plus the external load. Therefore the gain for CS amp is gm5*(ro5||ro6). Why we have a parallel connection ? Because when we are looking from the output terminal into the amplifier output we see two small-signal path for the current. One through ro5 and the second one through ro6.