I'm working on a problem where I need to determine whether the solution set $S$ to the equations $x^3 + y^3 +z^3 = 1$ and $xy = z$ in $\mathbb{R}$ is a smooth manifold.
I've defined $F: \mathbb{R}^3 \to \mathbb{R}^2$ as $(x,y,z) \mapsto (x^3+y^3+z^3,xy-z)$ and now the regular level set theorem implies that $S=F^{-1}(1,0)$ is a smooth submanifold if $(1,0)$ is a regular value.
Now $(1,0)$ is a regular value if $DF_{(x,y,z)}:T_{(x,y,z)}\mathbb{R}^3 \to T_{(1,0)}\mathbb{R}^2$ is surjective for all $(x,y,z) \in F^{-1}(1,0)$.
Computing the Jacobian I got that $$DF_{(x,y,z)} = \begin{pmatrix}3x^2&3y^2&3z^2\\ y&x&-1\end{pmatrix}.$$ However now linear algebra seems to get the best of me. I don't know how to compute the rank of this matrix. I've seen some argument based on the minors, but how does that even work for a non-square matrix?
Directly: $(1,0)$ fails to be a regular value if and only if all three $2\times 2$ minors are $0$. Note also that $x$ and $y$ appear symmetrically in this problem. So can we have \begin{align*} x^3-y^3 &= 0 \\ x^2+yz^2 &= 0 \\ y^2+xz^2 &= 0 \ ? \end{align*} As we expect, we get $x=y$, and so $x(x+z^2)=0$. So, either $x=y=0$ or $x=y=-z^2$. Now put these into your defining equations. What do you find?