Find the gain in mosfet
You could plot or write the expression for Vout vs Vin and then take the derivative of it to get the large signal gain \$ A_V= \frac{dVout}{dVin}\$ 
The simulation result is: 

 

For \$R_D=1k\Omega\$,\$R_S=100\Omega\$,\$V_{DD}=10V\$,\$V_{SS}=-10V\$

And notice the maximum gain is equal to \$\frac{R_D}{R_S} \approx 10\$

And this is not the coincident.  

Or you can do small signal analysis and solve for the gain.



From inspection we can write: 

$$V_{IN} = V_{GS}+I_D*R_S$$

And because \$I_D = g_mV_{GS}\$

We have 

$$V_{IN} = V_{GS}+g_mV_{GS}*R_S = V_{gs}\left (1+g_mR_S  \right ) $$

The output voltage can also be find by inspection 

$$V_{OUT} = -I_D*R_D$$

This minus sign comes from the fact that the \$I_D\$ current is flowing from GND into drain terminal in our  small signal equivalent circuit. 

In reality this minus sign only inform us that our amplifier output voltage is 180° out of phase with respect to the input voltage.

Any increases in \$Vin\$ will increase \$I_D\$ current also, the voltage across \$R_D\$ increases too, therefore \$V_D\$ voltage drops due to large voltage drop across\$R_D\$ resistor, \$V_d = V_{DD} - I_D*R_D\$    

Hence the voltage gain is equal to: 
$$\frac{V_{OUT}}{V_{IN}}= \frac{-g_mV_{gs}R_D}{V_{gs}\left (1+g_mR_S  \right )}= - \frac{g_mR_D}{1+g_mR_S} = - \frac{R_D}{\frac{1}{g_m} + R_S}$$