Then (y 3)' = 3y 2 * y'
(y^2)'=2y*y'
So y' * (3y 2+2y) = 2.
y'=2/(3y^2+2y)
Y= 1, so y'=2/5.
That is, tangent slope =2/5
y- 1=(2/5)(x- 1)
2x-5y+3=0