Let $S = \{ (x,y,z) \in \mathbb{R}^3 \mid x - yz + z^3 = 0 \}$. Let $\pi: \mathbb{R}^3 \to \mathbb{R}^2$ be such that $\pi(x,y,z) = (x,y)$.
Let $H = \{p \in S \mid \pi_{\mid S}: S \to \mathbb{R}^2 \text{ is not a local diffeomorphism in a neighborhood of } $p$ \}$. Where, $\pi_{\mid S}$ is $\pi$ restricted to the set $S$.
Show that $H$ is a smooth curve in $\mathbb{R}^3$ and find a parameterization for it.
Attempt: I know this is related to the multi-variate implicit or inverse function theorem, but I am not sure how to proceed next. In particular, I don't get what we should do with the local diffeomorphism condition in $H$.
We have $\frac{\partial f}{\partial x} = 1$. So the implicit function theorem tells us there is an implicit function such that $g(y,z) = x$. But I am not sure how to proceed from here.
Question: May I have a hint on how I should think about this generally, and how I should proceed next?
Note: This is not a homework problem, but I am hoping to solve it like one.
I actually don't know very much about differential geometry or topology, so bear that in mind.
My basic intuition and understanding of the question is this: we have a surface $S$ embedded in $\mathbb R^3$. If we project this surface onto the $xy$-plane by discarding the $z$ coordinate, some points of the surface will "flatten nicely," whereas some points will correspond to "folds" or "creases." It is these latter points that comprise the curve $H$.
According to what you wrote, you can say $g(y,z) = yz - z^3$. Then we want the critical points of $g$, namely $$y = 3z^2,$$ which in turn gives us $x = 2z^3$. We then have a parametrization of $H$ in terms of $t = z$: $$H = (x(t), y(t), z(t)) = (2t^3, 3t^2, t).$$ But this is not a proof and I have left out many details, because as I mentioned, I am not familiar with this topic.