xy^2 + 2xy = 8, then, at the point (1,2) y' is?

Answer 1

again, use implicit differentiation
y^2 + 2xyy' + 2y + 2xy' = 0
at (1,2) you have
4 + 4y' + 4 + 2y' = 0
y' = -4/3

Answer 2

Dear Gang of Bots,
The first reply by oobleck is correct.
Your replies are nonsense.