find the equation of the tangent line to the given curve at the indicated point

Question

x^2(y^2)=(y+1)^2 (4-y^2) at (0,-2)

Answer 1

2xy^2 y'+ 2x^2y y'= 2(y+1)y' (2-y^2)+ (y+1)^2(-2y y')

y' [( 2xy^2 +2x^2 y)-2(y+1)(2-y^2)+(y+1)^2 *2y]= 0

which means y' is zero, or the contents of [] is zero. So , at 0,-2

y' [0 + 0 -4]=0
If I evalauated that right, then y' has to be zero.

equation of tangent line
y=-2

check my work.