Implicit differentiation derivative not as expected

Question

I am trying to solve the problem: $x^{2}+xy+y^{3}=0$ using implicit differentiation.

My workings:

$(1)\,\,\,{d\over dx}{[x^2]}\,+{d\over dx}{[xy]}\,+\,{d\over dx}{[y^3]}={d\over dx}{[0]}$

$(2)\,\,\,2x+y+{{dy^3}\over{dy}}{{dy}\over{dx}}=0$

$(3)\,\,\,2x+y+3y^2\,({dy\over dx})=0$

$(4)\,\,\,{dy\over dx}=\boxed{-{{2x+y}\over{3y^2}}}$

But the answer says it should be:

$(3)\,\,\,2x+y+{{dxy}\over{dy}}{{dy}\over{dx}} + 3y^2\,({dy\over dx})=0$

$(4)\,\,\,2x+y+{dy\over dx}(x+3y^2)=0$

$(5)\,\,\,{dy\over dx}=-{2x\,+\,y\over{x\,+\,3y^2}}$

Why?

Answer 1

You can't conclude that $\frac{d}{dx}(xy)=y$, since this is effectively the product of the functions $x$ and $y$. Hence, using the product rule gives $$\frac{d}{dx}(xy)=y\frac{d}{dx}(x)+x\frac{d}{dx}(y)=y+x\frac{dy}{dx}$$ The rest should work out as expected.

edited to reflect Vegeta's comment

Answer 2

Differentials will make the process much more intuitive. $$\eqalign{ &0 = d\Big(x^2 + xy + y^3\Big) \\ &0 = 2x\,dx + \big(x\,dy+y\,dx\big) + 3y^2dy \\ &\big(x+3y^2\big)\,dy = -\big(2x+y\big)\,dx \\ &\frac{dy}{dx} = -\left(\frac{2x+y}{x+3y^2}\right) \\ }$$