I'm trying to solve the following problem:
For $c \in \mathbb{R} \setminus \{ 0 \}$, let $$C = \{(x,y) \mid x^3 + xy + y^3 = c \} \subset \mathbb{R}^2.$$
Show that for $c \neq 1/27$ the set $C$ is a closed one-dimensional submanifold of $\mathbb{R}^2$.
What I have so far
The following proposition seems very promising:
Proposition
If $f : M \rightarrow N$ has constant rank $k$ on a neighborhood of $f^{-1}(y)$, then $f^{-1}(y)$ is a closed submanifold of $M$ of dimesnsion $n - k$ (or is empty).
So I want to define my $f : \mathbb{R}^2 \rightarrow \mathbb{R}$ by $$f(x,y) = x^3 + xy + y^3 - c.$$
Now, $f^{-1}(0) = C$, and $\big[ \frac{\partial f}{\partial x} \ \ \ \ \frac{\partial f}{\partial y} \big] = \big[ 3x^2 + y \ \ \ \ \ 3y^2 + x \big]$.
But I'm not sure what to do from here. And why is $c = 1/27$ so special? Any help is appreciated!
Your map only fails to be a submersion when that matrix is zero. So we check when that happens.
This is precisely when $y = -3x^2$ and $x = -3y^2$, so $x = -27x^4$, or $x^3 = -1/27$, or $x = -1/3$. Then we must have $y = -1/3$ as well. So the map fails to be a submersion precisely at $(-1/3,-1/3)$; and this lies in $f^{-1}(0)$ for precisely one $c$: the $c$ such that $(-1/3)^3 +(-1/3)(-1/3) + (-1/3)^3 = c$; that is, for $c = 1/27$.
So unless $c = 1/27$, your map has constant rank (1) on $f^{-1}(0)$ as desired.