Let F(x,y,z)=x2y-2yz2+x2y-3=2x2y-2yz2-3=0 (I suspect there is a typo in the question. See later.)
∂F/∂x=4xy, ∂F/∂y=2x2-2z2, ∂F/∂z=-4yz.
So ∇F at (2,-3,1) is <-24,6,12>.
||∇F||=√((-24)2+62+122)=√(576+36+144)=6√21.
Unit normal vector, n=<-24,6,12>/(6√21)=<-4√21/21,√21/21,2√21/21>.
[I suspect that perhaps F(x,y,z)=x2y-2yz2+xy2-3.
∂F/∂x=2xy+y2, ∂F/∂y=x2-2z2+2xy, ∂F/∂z=-4yz.
So ∇F at (2,-3,1) is <-3,-10,12>.
||∇F||=√(9+100+144)=√253.
Unit normal vector, n=<-3√253/253,-10√253/253,12√253/253>.
I may not have the intended F(x,y,z), but I believe this is the correct method for finding the unit normal vector.]