I only dimly remember how to do this, so take this all with a grain of salt:
No, you can't assume that Vgs = 6V, because at DC Vs does not equal zero. Your source current will be in the neighborhood of 6V/2.7k-ohm, because while Vgs will be greater than zero, it won't be that much greater compared to 6V. If your data sheet gives you a means of finding gm from the source current, you're done.
The academic way to do this is to note that the source current is equal to some constant times (VT + Vgs)^2. Data sheets often give VT (IIRC it'll be called 'threshold' or 'pinch-off'), and they'll give the source current at Vgs = 0. This is enough to calculate the source current at any value of Vgs; you can solve the quadratic equation involving your source resistor to find the _exact_ Vgs, then you can differentiate the Is as a function of Vgs at that Vgs to find gm.
The academic way to do this leaves out a host of real-world issues, and is a PITA besides. Spice will tell you the gain of your circuit pretty quickly, and you can choke your gm value out of it without too much trouble, _and_ you can do things like Monte Carlo simulations to take into account the wide variation in JFET parameters while you're at it.
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Tim Wescott
Control systems and communications consulting
http://www.wescottdesign.com
Need to learn how to apply control theory in your embedded system?
"Applied Control Theory for Embedded Systems" by Tim Wescott
Elsevier/Newnes, http://www.wescottdesign.com/actfes/actfes.html
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