Implicit Function Theorem Application.

Question

I have the following problem

Consider the equation

$$x^3 + xy^2 + y^3 = 1$$

Is it possible get $x= x(y)$ at neighborhood of $(1,0)$? What about $y = y(x)$?

The first part, get $x=x(y)$ it's pretty easy, I just applied the theorem, but the second part I got stuck, because I can't conclude anything through the theorem, So I got a hint to try:

$$x^3 + xy^2 + y^3 = 1 \iff (x^3)' + x'y^2 + 2yx + 3y^2 = 0$$

But I don't know what conclude from above equation.

Thank you for any help.

Answer 1

For $y=y(x)$ to exist it is necessary for the Jacobian to be non-singular. In this case, the Jacobian is just a $1\times 1$ matrix. It is just a number with the value $$ \frac{d}{dy}(x^3 + xy^2 + y^3)=0+2xy+\frac{dx}{dy}y^2+3y^2=0 $$ at $y=0$. So the jacobian is sigular. So $y=y(x)$ dose not exist. If you try to sketch the curve, you will find it verticle at this point.

EDIT：Your question is kind of ambiguous. If you require $y=y(x)$ to be differentiable, then it does not exist. But you can certainly have an indifferentiable function $y(x)$ because a cubic always has a root.