Proof inverse f'n th'mimplicit f'n th'm?

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    Inverse Proof
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Show that if g: [0,1] x [0,1]-> R^3 is continuously differentiable with
det [Dg] (s',t') not equal to 0, then there exists S>0 such that
{g(s,t): (s,t) are elements of the ball, radius S, centred at (s',t')} is the graph of some function.
(s',t') is just some point.

I'm just not sure how to start. The prof said that it might be best to use inverse function theorem, but yah...any help would be great.
 
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what does the inverse function thm say
 
The inverse function theorem says:

Suppose U, a proper subset of R^n, is open, and x' is an element of U. F: U -> R^n is C1, and DF(x') is invertible. Then there is a neighbourhood V, a proper subset of U, of x' on which F has a C1 inverse function. So there are neighbourhoods V of x' and W of F(x'), and a function g: W-> V so that

F(g(y))=y for all y in W, and g(F(x))=x for all x in V.

Moreover, if F(x)=y, then

Dg(y)= (DF(x))^-1

[note: we're taking everything as vector-valued, although it should't really matter]
 
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