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?
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: $$\nabla g(1,1) = - \left( {\begin{array}{*{20}{c}} {\frac{3}{2}} \\ {\frac{5}{4}} \end{array}} \right)$$