Implicit Function! I need to show that equation  $z^{3} + z + xy=1$ defines an unique function on the set of real numbers  $g(x,y)=z$ ,for any x,y.Also i need to find $g'(1,1)$.This is what i have so far:
$F(x,y,z)=z^{3} + z + xy-1=0$,$F'_x=y$ and $F'_y=x$ are clearly continuous and $F'_z=3z^2+1\neq0$,for any real z.So by one of the theorem from my book,starting equation really defines an unique function $g=z(x,y)$,but i dont know how to find it.
 A: For the first part, proving the global existence of $g:\>{\mathbb R}^2\to{\mathbb R}$, the implicit function theorem is of no help. Argue as follows instead: The auxiliary function $$\phi:\quad{\mathbb R}\to{\mathbb R},\qquad z\mapsto w:=z^3+z$$ has a strictly positive derivative, so it is strictly increasing; furthermore one obviously has $\lim_{z\to\pm\infty}=\pm\infty$. It follows that $\phi$ maps ${\mathbb R}$ bijectively onto ${\mathbb R}$. Therefore $\phi$ has a continuously differentiable inverse
$$\psi:=\phi^{-1}:\quad{\mathbb R}\to{\mathbb R},\qquad w\mapsto z=\psi(w)$$
(but we don't have a simple formula for $\psi$). It is now easy to see that your $g$ is formally given by
$$g(x,y)=\psi(1-xy)\qquad\bigl((x,y)\in{\mathbb R}^2\bigr)\ ,\tag{1}$$
whence is $C^1$ as well.
In order to compute the partial derivatives of $g$ at the point $(x_0,y_0):=(1,1)$ we proceed as follows: Note that at $(1,1)$ we have $1-xy=0$; furthermore
$${\partial\over\partial x}(1-xy)=-y,\qquad {\partial\over\partial y}(1-xy)=-x\ .$$ Using the chain rule in $(1)$ we therefore obtain
$$g_x(1,1)=\psi'(0)(-1),\quad g_y(1,1)=\psi'(0)(-1)\ .\tag{2}$$
As $\phi(0)=0$ we have $\psi(0)=\phi^{-1}(0)=0$ and therefore, by a well known formula from one-variable calculus:
$$\psi'(0)={1\over\phi'\bigl(\psi(0)\bigr)}={1\over\phi'(0)}=1\ .$$
Plugging this into $(2)$ we obtain
$$\nabla g(1,1)=(-1,-1)\ .$$
