Series-Shunt Feedback Amplifier Analysis - Finding Loop Gain
Your circuit has voltage-series feedback (Series-Shunt). So, we can draw this equivalent circuit:







simulate this circuit &ndash; Schematic created using CircuitLab

The open-loop voltage gain \$A_{OL}\$ can be found out by inspection: 

\$M_1\$ voltage gain is: 

$$\frac{R_{D1}}{R_{S1}||R_F + \frac{1}{g_{m1}}} \approx 29.41 $$ 

\$M_2\$ voltage gain:

$$\frac{R_{D1}}{\frac{1}{g_{m2}}}=\frac{10k\Omega}{250 \Omega} = 40 $$ 

And \$M_3\$ voltage gain: 

$$\frac{\left ( R_{F}+R_{S1} \right )||R_{S2}}{\frac{1}{g_{m3}}+\left ( R_{F}+R_{S1} \right )||R_{S2}} \approx 0.267 $$ 

Hence the open-loop voltage gain is :

$$A_{OL} \approx 314 $$

The feedback factor 

$$\beta = \frac{V_{S1}}{V_O} = \frac{R_{S1}}{R_{S1}+R_F} = 0.1 $$ 

And finally the loop gain: 

$$A_{OL}\,\beta = 314*0.1 = 31.4$$   

And the Closed loop gain 

$$A_{CL} = \frac{A_{OL}}{1+A_{OL}\,\beta} \approx 9.69$$