Consider the following subsets of $\mathbb{R^3}$
$1$) $S_1 = \{ (x,y,z) \in \mathbb{R^3} : x^3+y^3+z^3 = 1\}$
$2$) $S_2 = \{(x,y,z) \in \mathbb{R^3} : x^3+y^3+z^3 = 1 , x+y+z=0 \}$
I need to show that $S_1$ and $S_2$ are regular submanifolds of $\mathbb{R^3}$.
So for $S_1$, I calculated the critical points which can be found either by writing a Jacobian or simply differentiating the function and equating it to $0$. The only critical point is $(0,0,0)$ and since this point doesn't belong to $S_1$, s0 every point of $S_1$ is a regular point and hence $S_1$ is a regular submanifold. Is my approach correct?
However, if I apply the same reasoning for $S_2$, I run into trouble as here the critical point $(0,0,0)$ does lie in $S_2$ and so it can not be a regular submanifold as every point is not a regular point. Am I wrong here?
If I am not right then please help me with the solution!
For $S_2$, you know that each piece is a submanifold ($x+y+z = 0$ is a plane). Now just show that the intersection of the manifolds (which are the solution sets of) $x^3+y^3+z^3=1,x+y+z = 0$ is transverse and you are done.