Show that $\frac{\partial g}{\partial x_1}(a_1,a_2)=-\frac{\partial f/\partial x_1}{\partial f/\partial y}(a_1,a_2,b)$ The implicit-function theorem implies that, for points $(x, y)$ near $(a, b)$, the equation $f(x, y) = f(a, b)$ determines $y$ as a $C^r$ function $y = g(x)$.
Suppose that $X = \mathbb{R}^2$ and that $Y = W = \mathbb{R}$. Thus $(x,y) = (x_1,x_2,y)$, and the hypothesis is that $(\partial f/ \partial y)_{(a_1,a_2,b)} \ne 0$. 
Show that $\frac{\partial g}{\partial x_1}(a_1,a_2)=-\frac{\partial f/\partial x_1}{\partial f/\partial y}(a_1,a_2,b)$ and $\frac{\partial g}{\partial x_2}(a_1,a_2)=-\frac{\partial f/\partial x_2}{\partial f/\partial y}(a_1,a_2,b)$.
I am very confused by the implicit function theorem. I'm sure there is a way to show directly that $\frac{\partial g}{\partial x_1}(a_1,a_2)=-\frac{\partial f/\partial x_1}{\partial f/\partial y}(a_1,a_2,b)$ and that the other will be nearly the same. But I am not sure what $\frac{\partial g}{\partial x_1}(a_1,a_2)$ really means. I know that it is the partial derivative with respect to $x_1$, but how can I evaluate it without a definition of the function?
 A: You have to derivate $f(x_1,x_2,g(x_1,x_2))$ and see what you can do.
$$\frac{df}{dx_1}=\frac{\partial f}{\partial x_1}+\frac{\partial g}{\partial x_1}\frac{\partial f}{\partial g}=\frac{\partial f}{\partial x_1}+\frac{\partial g}{\partial x_1}\frac{\partial f}{\partial y}$$
This formula is true while $f(x_1,x_2,g(x_1,x_2)) \equiv 0$, so $\frac{df}{dx_1}=0$. Hence :
$$\frac{\partial g}{\partial x_1}=-\frac{\partial f/\partial x_1}{\partial f/\partial y}$$ 
The same can be done for $\frac{df}{dx_2}$.
By the way : $\frac{\partial g}{\partial x_1}(a_1,a_2)$ means what you think it means, you take the function $g$, you take the partial derivative and you evaluate it on $(a_1,a_2)$. The fact that $g$ is not necessarily known is not a problem for considering this quantity, which is, as you can see, totally determined by $f$. (Furthermore, this is very useful to know $\frac{\partial g}{\partial x}$  in a lot of problems)
A: The implicit function theorem states a few things. Apart from determining the derivate of the implicit function it also guarantees the unique existence of such a function (it doesn't however guarantee that there's a closed form expression for that function). 
Before you prove the derivate you have to at least prove the existence of the implicit function. The proof of that you use the fact that since it has a gradient and it's continuous you can use the fact that it's monotone along the $y$ axis in an neighborhood and also that there's higher and lower values, this means that the equation $f(x,y) = f(a,b)$ has an unique solution for each $x$ in a neighborhood $a$. We therefore have a unique function such that $f(x, g(x)) = f(a,b)$.
Next step is to ensure that $g$ is differentiable. To do this we use the differentiability of $f$ to get an estimate of $g(x)$ that is enough to show it's differentiability. Recall that
$$f(x+h_x, y+h_y) = f(x,y) + \nabla f(x,y)\cdot(h_x, h_y) + o(h_x, h_y)$$
By inserting $g$ we get:
$$f(x+h, g(x+h)) = f(x,y) + \nabla f(x,y)\cdot(h, g(x+h)-g(x)) + o(h, g(x+h)-g(x))$$
what we need here is to ensure that $g(x+h)-g(x)$ behaves well enough. We can do this by putting an estimate on it, if we for example know that there's an $K$ such that $|g(x+h)-g(x)|<Kh$ we get that the ordo is $o(h)$, and then we can solve for $g(x+h)-g(x)$ above and use that to calculate the derivate. That $|g(x+h)-g(x)|<Kh$ follows can be shown by using the estimate for $f(x+h_x, y+h_y)$.
(actually the second step we only need to show that $g$ is differentiable, because then the chain rule will require it's derivate to be according to the implicit function theorem).
