Implicit function theorem - how to approach? I have a question that I have been working on for a while. I was wondering how I should approach the following question:

Are there any points on the graph of the equation 
  $$x^3+3xy^2+2xy^3=1$$
  where it may not be possible to solve for $y$ as a smooth function of $x$ in
  some neighborhood of the point?

 A: You could calculate the Jacobian matrix and determine which points make the column corresponding to the partial derivative of $y$ pivotal/nonpivotal.
EDIT:
The derivative matrix would be $\begin{bmatrix}
3x^2+3y^2, & 6xy+6xy^2
\end{bmatrix}$. What points make $3x^2+3y^2 = 0$ and $6xy+6xy^2\neq 0$? These would be the points where you can solve for $y$ in terms of $x$. 
A: Another way to do this kind of problem is to use the implicit derivative. Anywhere on the curve where the implicit derivative $dy/dx$ is $\pm \infty$ (on such simple curves), the curve will be momentarily vertical, and at those points $y$ cannot locally be a smooth function of $x$. For your function $f(x,y)=x^3+3xy^2+2xy^3-1$, so curve is $f(x,y)=0$, the implicit derivative is
$$\frac{dy}{dx}=-\frac{2y^3+3y^2+3x^2}{6xy(y+1)}.$$
So $dy/dx$ will be infinite at points $(x,y)$ of the curve at which the denominator is $0$ and the numerator nonzero. Here the denominator is $0$ at $x=0$, however no points on $f(x,y)=0$ have $x=0$. So there remain the possibilities $y=0,-1$. For $y=0$, substitution in $f(x,y)=0$ gives $x=1$, so that one of the "bad" points is $(1,0).$ And for $y=-1$, substitution in $f(x,y)=0$ gives the equation $x^3+x-1=0$ which has a single solution around $x=0.682328.$ This gives the other "bad" point as $(0.682328,-1)$ approximately. At all other points on the curve, $dy/dx$ is a finite number and so the implicit function theorem says $y$ is locally a smooth function of $x$ near the point.
