I want to verify that the equation $x^3+xy^2+y^3=1$ can be locally given at $(1,0)$ by $x=x(y)$ and by $y=y(x)$.

For this, I thought in use the implicit function theorem. I defined $\displaystyle f:\mathbb{R}^2 \rightarrow \mathbb{R}$ as $\displaystyle f(x,y)=x^3+xy^2+y^3$. So, $\displaystyle \nabla f(x,y)=(3x^2+y^2,3y^2+2xy)$ and then $\displaystyle \nabla f(1,0)=(3,0)$.

Since $\displaystyle \dfrac{\partial f}{\partial x}(1,0)=3\neq 0$, we have that the equation can be given by $x=x(y)$. However, $\displaystyle \dfrac{\partial f}{\partial y}(1,0)=0$ and I don't know how to decide if the equation can be given by $y=y(x)$ (the theorem doesn't work here).

How do I proceed? Can someone help me? Thank you!


This is a picture representing the points $(x,y)\in[0,2]\times[-1,1]$ such that $x^3+x y^2+y^3=1$:

enter image description here

This curve is in a neighbourhood of $(1,0)$ the graph of a $x=x(y)$ function, but not the graph of a $y=y(x)$ function. You may just show that for any $x=1-\varepsilon$ with $\varepsilon>0$, there are two distinct values of $y$, a negative one and a positive one, such that $x^3+xy^2+y^3=1$ holds, since $g(y)=y^3+y^2$ has a local minimum at $y=0$.

  • $\begingroup$ But, is there a mathematical way to conclude it without see this picture? $\endgroup$ – Rodrigo Dias Oct 20 '16 at 16:17
  • $\begingroup$ @rdias: just added. $\endgroup$ – Jack D'Aurizio Oct 20 '16 at 16:18
  • $\begingroup$ So, in general, it's hard to conclude it, right? Already in this equation, that is relatively simple, it's hard to do it, who knows about other more comṕlex equations?! $\endgroup$ – Rodrigo Dias Oct 20 '16 at 16:20
  • 1
    $\begingroup$ @rdias: if some partial derivative vanish, we have to consider more terms in a local Taylor expansion. In this case we may conclude what stated above from the fact that $g(y)=y^3+y^2$ has a local minimum at $y=0$, but in general it is not trivial. $\endgroup$ – Jack D'Aurizio Oct 20 '16 at 16:23
  • $\begingroup$ For instance, $x^2+xy^3+y^5=1$ is in a neighbourhood of $(1,0)$ both the graph of a $x=x(y)$ and a $y=y(x)$ function, even if $\frac{\partial f}{\partial y}=0$. $\endgroup$ – Jack D'Aurizio Oct 20 '16 at 16:28

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