minimize F=x^2+y^2 with the constraint xy^2=16  show work

Hello and thank you for your question.

F = x^2 + y^2
and
xy^2 = 16 or y^2 = 16/x

F=x^2 + 16x^(-1)

finding the derivative:
F' = 2x - 16x^(-2)

solving for F' = 0
0 = 2x - 16x^(-2)
factoring out x
0 = x  [but x=0 is not a solution because of division by 0 in y^2 = 16/x ]
0 = 2 - 16x^(-3)  
2 = 16x^(-3)  
1/8 = x^(-3)  
  x = 2
  F(2)= 4 + 8 = 12
So (2,12) is the minimum of F because that's when F' = 0

Thanks you agreed with what I kept on coming up with and the teacher
had something entirely different