Let $M = \{(x, y, z) \in {\rm I\!R}^3 : F(x,y,z) = 0\}$ and let $F(x,y,z) = (3x^2z + y^2 + z^3-1, \, x + z-1)$ .
Does the function $f(x, y, z) = x$ have any extrema in $M$?
We are asked in advance to prove that $M$ is a manifold. Thus I believe that we have to use the method of Lagrangian Multipliers to solve the exercise, but I can't seem to solve the system.
This is what I have so far:
\begin{cases} (1,0,0) = \lambda_{1}(6xz,\,2y,\,3z^2) + \lambda_{2}(1,\,0,1) \\3x^2z + y^2 + z^3=1 \\ x+z = 1 \end{cases}