Intuitive Understanding About the Implicit Function Theorem In multivariable calculus, given a function like $F(x,y,z) =0$, the implicit function $z=f(x,y)$ exists if and only if $\frac{\partial F}{\partial z} \not = 0$. And the implicit function is given by $\frac{dz}{dy}=-\frac{\frac{\partial F}{\partial y}}{\frac{\partial F}{\partial z}}$. 
The theorem can be proved by mathematical reasoning. But I want to know whether there is some intuitive understanding of the theorem, say geometry intuition in the three dimensional space, etc.
 A: If you question ask for the intuition concerning the existence of the implicit function, let me suggest the following. 
I prefer to simplify by restricting to two variables. The theorem states that if at $(x_0,y_0)$ the curve  $\{(x,y)\in\mathbb R^2 : F(x,y)=0\}$ does not look  like a vertical line (i.e if $\partial F/\partial y(x_0,y_0)\neq 0$) then the curve is locally the graph of a function $x\mapsto y(x)$. A good example to have in mind is the circle, $F(x,y)=x^2+y^2-1$. The same reasoning works for higher dimensions. 
If you ask about the form of the implicit function, the answer of copper.hat looks nigh enough to me.
A: Essentially, the Inverse Function Theorem tells you that if a continuously differentiable function is has an invertible derivative at a point (i.e. the linearization of the function there has an inverse) then the function itself is invertible on some neighborhood of the point. The Implicit Function Theorem is just a special case of the Inverse Function Theorem, which follows from defining $G(x,y,z)=(z,F(x,y,z))$.
A: First, the IFT is a sufficient condition; if you let $F(x,y) = x-y^3$, then an implicit function exists ($y = \sqrt[3]{x}$), but $\frac{\partial F(x,0)}{\partial y} = 0$.
Regarding intuition, think in terms of linear functions. Just take $F(x,y) = a x + b y$. Notice that $\frac{\partial F(x,y)}{\partial x} = a$, $\frac{\partial F(x,y)}{\partial y} = b$. If $b\neq 0$, then the implicit function is explicitly given by $ y = -\frac{a}{b} x$, from which you can see that $\frac{\partial y(x)}{\partial x} = -\frac{a}{b} = - (\frac{\partial F(x,y)}{\partial y})^{-1} \frac{\partial F(x,y)}{\partial x}$.
