<s>[INST] Application of Implicit Function Theorem for Constrained Optimization Here's the problem:
Consider the subset $S \subset \mathbb R$ defined by 
$$ x^4+2xy+y^4+yz+z^3 = 2 $$
Show that there exists a $C^1$ function $g: \mathbb R^2 \to \mathbb R$ defined near $(1,1)$ such that for $(x,y,z)$ near $(1,1,-1)$, then $(x,y,z) \in S $ if and only if $z=g(x,y)$. Compute $\nabla g $ near $(1,1)$.
Here's my attempt at a solution:
I set the equation equal to $0$ and computed $-\frac{F_x}{F_z}$ and got 
$$\left. \frac{-4x^3+2y}{y+3z^2}\right|_{(1,1,-1)}=-\frac{1}{2}$$
So, can I say that, because this does not equal $0$, such a function exists? If so, how can I find that function so I can compute its gradient?
 [/INST] From the equation:
$${x^4} + 2xy + {y^4} + yz + {z^3} = 2$$
We define a function:
$$F(x,y,z)= {x^4} + 2xy + {y^4} + yz + {z^3} - 2$$
Equation reads now:
$$F(x,y,z) = 0$$
Because of
$$dF(x,y,z) = (4{x^3} + 2y)dx + (2x + 4{y^3} + z)dy + (y + 3{z^2})dz$$
we have:
$$\nabla F(x,y,z) = \left( {\begin{array}{*{20}{c}}
  {4{x^3} + 2y} \\ 
  {2x + 4{y^3} + z} \\ 
  {y + 3{z^2}} 
\end{array}} \right)$$
So $$\frac{{\partial F}}{{\partial z}}(x,y,z) = y + 3{z^2}$$
and
$$\frac{{\partial F}}{{\partial z}}(1,1, - 1) = 1 + 3 = 4 \ne 0$$
Because of:
$$\frac{{\partial F}}{{\partial z}}(1,1, - 1) \ne 0$$
near $$(1,1, - 1)$$ exists a function $$z = g(x,y)$$ with
$$F(x,y,g(x,y)) \equiv 0$$
near $$(1,1)$$ and
$$\begin{gathered}
  (x,y,z) = (x,y,g(x,y)) \hfill \\
  (1,1, - 1) = (1,1,g(1,1)) \hfill \\
  g(1,1) =  - 1 \hfill \\ 
\end{gathered} $$
Setting
$$(4{x^3} + 2y)dx + (2x + 4{y^3} + z)dy + (y + 3{z^2})dz = 0$$
it follows:
$$dz(x,y) =  - \frac{{4{x^3} + 2y}}{{y + 3{z^2}}}dx - \frac{{2x + 4{y^3} + z}}{{y + 3{z^2}}}dy$$
with:
$$\begin{gathered}
  dz(1,1) =  - \frac{6}{{1 + 3g{{(1,1)}^2}}}dx - \frac{{2x + 4{y^3} + g(1,1)}}{{1 + 3g{{(1,1)}^2}}}dy \hfill \\
  dz(1,1) =  - \frac{3}{2}dx - \frac{5}{4}dy \hfill \\ 
\end{gathered} $$
That means:
$$\nabla g(1,1) =  - \left( {\begin{array}{*{20}{c}}
  {\frac{3}{2}} \\ 
  {\frac{5}{4}} 
\end{array}} \right)$$
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