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