Determining the equivalent resistance
EDIT: it has been a while since I looked at this kind of circuit problem, so I think I made a mistake in defining \$V_{gs}\$, but it doesn't matter, and here's why:

Firstly, there can be no current in the loop that has \$R_{s}\$ and \$R_{g}\$ in it.

As for the current source: it is in parallel with \$V\$ and \$R_{DS}\$, so if you do have any current from your current supply, it has to go through the voltage supply and can't go through the resistor \$R_{DS}\$ or the voltage across the resistor would be higher than the voltage supply in parallel with it \$V\$ and that breaks all the rules.  If you need to verify this with some equations/analysis, do mesh analysis on the right half of the circuit and verify that the current supply can't put any current in \$R_{DS}\$.

So you're left with a voltage supply (\$V\$) and a resistor (\$R_{DS}\$) in parallel.  The only current that can go through the resistor has to come from the voltage supply.  

To get the equivalent resistance \$R = V/I\$, first rewrite:

\$g_{m}*V_{gs} = -g_{m}*V\$, 

since node D and G are the same, and \$V_{DS}\$ is equivalent to \$-V\$.

Next, write out \$I\$ as the sum of the negative current from the current source plus the current from \$V/R_{DS}\$:

\$I = g_{m}*V + V/R_{DS} \$

Next, substitute our derived \$I\$ into the equation \$R=V/I\$ and simplify to get the answer:

\$R = V/I = V/( g_{m}*V + V/R_{DS}) = 1/(g_{m} + 1/R_{DS})\$

And that is the equivalent resistance.