The circuit is an op-amp with ground as the negative rail. This means Vin cannot go below the negative rail. Now look at the output current formula - it involves Vin and, if Vin cannot go below ground, then the current through the MOSFET HAS to be positive only.
In short, you cannot put an AC voltage at Vin unless it is biased to make its most negative excursion higher than ground and this of course means you CANNOT get an AC load current through the MOSFET. At best it will be an AC load current superimposed on a DC load current whose most negative current is still slightly positive with respect to the Iout arrow in the picture.
Another thing that trouble me is that why Iout only depend on that 1
Ohm resistor and not RL?
Because the op-amp sets the voltage across the 1 ohm resistor by appropriately driving the MOSFET to equal Vin. This means that Iout MUST equal Vin/1ohm. This is bread-and-butter for an op-amp operating linearly.