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 $x^3+x y^2+y^3=1$:
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 $y=0$.