# The converse of the Implicit Theorem

According to the implicit function theorem(on $$\mathbb R^2$$ for simplicity), if $$\displaystyle\frac{\partial f}{\partial y}\ne 0$$ at $$(x_0, y_0)$$, then on a neighborhood of $$(x_0, y_0)$$, there is a function $$g\in C^1$$ such that $$y = g(x)$$

There was no information about the converse in the text book, but one day I wondered that they is a function $$g\in C^1$$ such that $$f(x, g(x)) = 0$$ even though $$\displaystyle\frac{\partial f}{\partial y}= 0$$ ? Because $$\displaystyle\frac{\partial f}{\partial y}$$ takes part in deonminator, $$g$$ won't have a derivative function(or maybe another closed-form), and I have found $$g$$ but the only continuity works. Can we find the example that the implicit function $$g$$ exists even though the partial derivative at that point is zero? If no, how can I prove this, and which condition can be added to make the Implicit function theorem an equivalent conditions?

Take $$f(x,y)=(y-x^2)^2$$ and any point $$(x_0,y_0)$$ on the curve $$y=x^2$$. Then $$\nabla f(x_0,y_0) = (0,0)$$, so the implicit function theorem doesn't apply, but still the equation $$f=0$$ defines a nice function, namely $$y=g(x)=x^2$$.
But if $$f \in C^1$$ and $$f_y(x_0,y_0)=0$$ and $$f_x(x_0,y_0)\neq 0$$, and the equation $$f=0$$ happens to define a function $$y=g(x)$$ implicitly near $$(x_0,y_0)$$, then that function $$g$$ can't be differentiable at $$x_0$$, for precisely the reason that you give; namely, in that case implicit differentiation would give $$0=f'_x(x_0,y_0) + f'_y(x_0,y_0) g'(x_0)$$, which is incompatible with the assumptions.
Another example which might interest you is $$f(x) = \begin{cases} \dfrac{y}{\sqrt{x^2+y^2}}, & (x,y) \neq (0,0), \\ 0, & (x,y) = (0,0), \end{cases}$$ which isn't even continuous at the origin, and hence not $$C^1$$ either. But still the equation $$f=0$$ defines a $$C^\infty$$ function $$y=g(x)=0$$.
Arguably the simplest, most devastating counterexample is $$f(x, y) = y^{3}$$, whose zero level is the $$x$$-axis, the graph of a constant function, but whose differential is identically zero on the entire level curve.
(Let's not nitpick about whether $$f(x, y) = y^{2}$$ is simpler. Cubing is a smooth bijection of the reals while squaring isn't, so we avoid even the whiff of certain spurious conjectures.)