The question is: define $f$ in $R^2$ by $f(x,y)=2x^3-3x^2+2y^3+3y^2$.
A) Find the four points in $R^2$ at which the gradient of $f$ is zero. Show that $f$ has exactly one local maximum and minimum in $R^2$.
B) Let $S$ be the set of all $(x,y)\in R^2$ at which $f(x,y)=0$. Find those points of $S$ that have no neighborhoods in which the equation $f(x,y)=0$ can be solved for $y$ in terms of $x$ (or $x$ in terms of $y$). Describe $S$ as precisely as you can.
So part A is easy to do and you can easily show $(1,0)$ is the only local minimum, and $(0,-1)$ is the only local maximum. The points $(0,0)$, and $(1,-1)$ are saddle points. I have done the algebra to solve when $f(x,y)=0$, which can be shown at $$x=-y$$, and $$y={2x-3\pm\sqrt{9+12x-12x^2}\over 4}$$ You can solve for the quadratic solution of $x$ with a simple substitution. My Question now comes about showing that $(0,0)$, and $(1,-1)$ are the only two points where it isn't analytically solvable for $y$ in terms of $x$ (or $x$ in terms of $y$). Is it just a coincidence that those are the only two saddle points or is there something deeper about the Hessian matrix being negative there where the gradient is zero? I don't think the Jacobian tells you anything since it is a $2\times1$ vector. I know the question is getting at the necessary conditions for the implicit function theorem, which I am struggling to understand regardless.