Electronic – Self-Biased Common-Source Stage

No, you are wrong.

You wrote: "...This in turn decreases the drain current of M2 causing a decrease in the voltage at point x.".

The right statement would be:

This in turn increases the M2 drain-source channel resistance causing an increase of the voltage at point x. (This voltage, \$ V_x = I_{in} \cdot R_{channel} \$, while both \$ I_{in} \$ and \$ R_{channel} \$ are increasing in this case).

Don't forget the input source is a current source, not a voltage one. The only path for its current is the drain of M2 -> ground, no current can flow into the M1 gate. When you increase the input current, it can't decrease as it is an independent current source (it is under your full control). So your claim: ...This in turn decreases the drain current of M2... is a contradiction.

The feedback is positive.

A supply of 10 volts across 1k means 10mA is the maximum current you can get thru the MOSFET. That is the problem with your experiment - no matter how much more you turned the juice up on the gate, the aiming point for drain current is 10mA and can not be exceded no matter how hard you try.

Of course Vds decreased - the mosfet was turning really hard-on and acting as a very small value resistor. This small resistance forms a potential divider with the 1k to make a very small voltage that is nowhere near 10V.