Find equations of the tangent and normal lines to the curve (x + y2)3 = 6x2y2 + 3 at the point (2, -1)?

Answer 1

Derivate both sides
3(x+y^2)^2*(1+2yy')=6( 2xy^2+2x^2yy')
Let´s substitute x andy
27(1-2y') =12(2-4y') 27 -54y'=24 -48y' 6y'=3 and y'=1/2
So the tangent is
y+1=1/2(x-2) and the normal is
y+1=-2(x-2)