implicit differentiation

Question

Find the slope of the tangent line to the curve xy^3 + 2y - 0.76 =0 at the point (-3, o.7)

I'm lost.
so 3xy^2 * y' + 2y = 0 right?

Answer 1

not quite.

xy^3 + 2y - 0.76 =0
y^3 + 3xy^2 y' + 2y' = 0
y' = -y^3/(3xy^2 + 2)

xy^3 is a product f*g
(f*g(' = f'g + fg'

at (-3,0.7) y' = -.7^3/(3(-3)(.7^2) + 2)
= . . .