Condition for Implicit Function Theorem Given $f(x,y)$ continuously differentiable in a domain of the plane, $f(a,b) = 0$ and $f_x(a,b)\ne 0$. Can you show $f(x,y) = 0$ in a finite neighborhood of $(a,b)$?

Examples

Sphere tangent to the plane.
$f(a,b )= 0$ at one point but $f_x(a,b) = 0$ and $f_y(a,b) = 0$ and there is no neighborhood of $(a,b)$ in which $f(x,y)=0$.

Sphere intersecting the plane.
There are points $(a,b)$ (on the circle of intersection) where $f(a,b) =0 $ , $f_x(a,b)\ne 0$ , and $f(x,y)=0$ in a neighborhood of $(a,b)$, (circular arc through $(a,b)$ )


But how do you show this in general without geometry?
Application: Implicit Function Theorem.
Utterly false:
$$f(x,y) = x,\qquad f(0,0) = 0,\qquad f_x(0,0) = 1\ne 0.$$
Inverse Function Theorem

Given E a subset of Rn and f:E->Rn  

If y=f(x), if Difj=dfj/dxi i,j=1,2,3, exist and are continuous on E, and if det (Difj) =’0, at x=a, and if b=f(a)
Then y=f(x) is invertible in a finite neighborhood of y=b.  

Illustration of terminology:  

y1=f1(x1,x2,x3), b1=f1(a1,a2,a3)
y2=f2(x1,x2,x3), b2=f2(a1,a2,a3)
y3=f3(x1,x2,x3), b3=f3(a1,a2,a3)    

x1=g1(y1,y2,y3)
x2=g2(y1,y2,y3)
x3=g3(y1,y2,y3)

Finite Neighborhood: not just in the limit as (y1,y2,y3)->(b1,b2,b3)

Implicit Function Theorem

Let z=f(x,y)
z1=f1(x1,x2,y)
z2=f2(x1,x2,y)  

Let Difj  (D3f1=df1/dy, D3f2=df2/dy) i=1,2,3 j=1,2 exist and be continuous.
Let det (D1f1,D1f2; D2f1,D2f2)=’0 at x,y=a,b where f(a,b)=0.

Consider the map F(z,w) where z=f(x,y), w=y
F1=f1(x1,x2)
F2=f2(x1,x2)
F3=y  

Then Difj i,j=1,2,3 exists and is continuous. Furthermore det (DiFj)=’0. By the Inverse Function Theorem it follows that F is invertible in a neighborhood of (f(a,b),b)=(0,b):  

x1=g1(F1,F2,F3)
x2=g2(F1,F2,F3)
y=y  

So far so good. If F1=0 and F2=0 in a subset of the neighborhood of (f(a,b)=0,b), then  

x1=g1(0,0,y)
x2=g2(0,0,y)  

in a neighborhood of y=b, ie, x=g(y).
But you can’t assume this. You have to prove F1=0 and F2=0 in a subset of  a neighborhood of (0,b) under given conditions, ie, f(x,y)=0 in a subset of a neighborhood of x,y=a,b. I haven’t seen this proof anywhere. 

Conclusion: The Implicit Function theorem is unproven, ie, you can’t start off a discussion of implicit functions by writing;
f1(x1,x2,x3,y1,y2)=0
f2(x1,x2,x3,y1,y2)=0
f3(x1,x2,x3,y1,y2)=0
because you don’t know that condition exists, unless you solve it in which case the Implicit Function Theorem is academic. Assuming it exists is tantamount to assuming it is invertible (under usual conditions).

Ref: Baby Rudin, Inverse and Implicit function theorem. Also 3rd Ed.