Trying to prove M is a manifold Let $M$ be the set of all points $(x, y, z) \in \mathbb{R^3}$ satisfying both of the equations
$x^3 + y^3 + z^3 = 1$ and $x + y + z = 1$. Prove that M is a manifold, except
perhaps near the points $(x; y; z) = (-1; 1; 1) ; (1;-1; 1) ; (1; 1;-1)$. What is
the dimension of $M$? 
I have calculated my $D_m=(3x^2 + 3y^2 + 3z^2)$ but I am not sure where to go from here.

Consider $\Phi = x^3 + y^3 + z^3 - 1$, and $\Psi = x + y + z - 1$, and $\Lambda = (\Phi, \Psi)$, then $M = \Lambda^{-1}(0,0)$.
Consider the matrix $\begin{pmatrix} \Phi_x & \Phi_y & \Phi_z \\ \Psi_x & \Psi_y & \Psi_z \end{pmatrix} = \begin{pmatrix} 3x^2 & 3y^2 & 3z^2 \\ 1 & 1 & 1 \end{pmatrix}$
Consider the matrices : $D_1 = \begin{pmatrix} 3x^2 & 3y^2 \\ 1 & 1 \end{pmatrix}$.
$D_2 = \begin{pmatrix} 3x^2 & 3z^2 \\ 1 & 1 \end{pmatrix}$
$D_3 = \begin{pmatrix} 3y^2 & 3z^2 \\ 1 & 1 \end{pmatrix}$
We have: $\det(D_1) = 3(x^2 - y^2)$
$\det(D_2) = 3(x^2 - z^2)$
$\det(D_3) = 3(y^2 - z^2)$.
To prove that $M$ is a manifold we only need to prove that it cannot occur at the same time that: $\det(D_1) = \det(D_2) = \det(D_3) = 0$. So suppose it can. Then this means:
$x^2 = y^2 = z^2$.
So if $x = y = z$, then $x = y = z = \dfrac{1}{3}$, and $x^3 = y^3 = z^3 = \dfrac{1}{3}$, contradiction. 
So if $x = -y$ then $x^3 + y^3 = 0 = x + y$. So $z = z^3 = 1$, but then $y^2 = z^2 = 1$, so $y = 1$ or $-1$. Then $x = 1$ or $-1$. But these points are excluded in the problem so it cannot be the case that all three determinants be zero. Thus one of them must be non-zero, which means the matrix has full rank = $2$ which is also the dimension of $M$.
