find unit normal vector to the surface

Question

find the unit normal vector  at (2,-3,1) to the surface x^2y-2yz^2+x^2y-3=0

Answer 1

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.]

Answer 2

_​​​x²y-2xz+2y²z⁴=10.  at.    (2,1,-1)