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?
Really appreciate your helps.
 A: 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 $y=g(x)=0$.
A: tl; dr: There is No Hope of a converse along the suggested lines.

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.)
