I find Implicit function theorem (limited to scalar function) both fascinating and a bit puzzling. Simply said, total derivative of some $f(x,y(x))$ according to $x$ is:
$$ \frac{d f(x,y(x)) }{ dx } = f_{x} \frac{\partial x}{\partial x} + f_{y} \frac{\partial y}{\partial x} = 0$$
It is straight forward to see, that such total derivative of $f(x,y(x))$ can be expressed through partial derivative:
$$\frac{\partial y}{\partial x} = - \frac{f_{x}(x, y(x))}{f_{y}(x,y(x))} \tag {1}$$
where for brevity $f_{x} = \frac{\partial f(x,y(y))}{\partial x}$ and $f_{y} = \frac{\partial f(x,y(y))}{\partial y}$
So far, so good. Now lets try to continue by taking partial derivative with respect to $y$. This can naturally happen when one needs to evaluate implicit function $f(x,y(x))$ for variable $y$ at variable $x$, it's first derivative $\frac{\partial y}{\partial x}$ and naturally one would expect second derivative $\frac{\partial ^2 y}{\partial ^2 x}$ and $\frac{\partial ^2 y}{\partial x \partial y}$ to exist, e.g. for calculating Hessian matrix.
Case 1: apply quotient rule on {1}
One could blindly do the following in order to obtain $\frac{\partial}{\partial y}\frac{\partial y}{\partial x}$ to get $\frac{\partial ^2 y}{\partial x \partial y}$:
$$\frac{\partial ^2 y}{\partial x \partial y} = - \frac{\partial (\frac{f_{x}}{f_{y}})}{\partial y}$$ which holds, if higher derivatives of $f_x$ and $f_y$ exists.
Case 2: analyze where the expression becomes constant
Obviously $\frac{\partial y}{\partial y} = 1$. This happens within $\frac{\partial ^2 y}{\partial y \partial x}$ and since $\frac{\partial ^2 y}{\partial x \partial y} = \frac{\partial ^2 y}{\partial y \partial x}$, then it is expected to happen for any mixed second-order partial derivative.
This way, one should have always $\frac{\partial ^2 y}{\partial y \partial x} = 0$ for an implicit function, since deriving $\frac{\partial y}{\partial y} = 1$ is the derivation of a constant, thus 0.
Question
Which of the 2 cases is wrong and why?
More details
To provide more details, but still keep the question general, lets assume the form of $f(x,y(x))$, to provide some visual clues.
$$f(x,y(x)) = e^{(x+y)} + x + y = 0$$
Background (to give the equations intuitive motivation related to experimental observations)
Variables $x$ and $y$ are physical quantities (current, voltage). $f(x,y(x)) = 0$, thus this function is used to evaluate the quantities $x$ and $y$ only. For instance, I need the value of $y$ for $x=c$, then I evaluate $f(c,y) = 0$ to get $y$. This also means, that when one variable changes (e.g. $x$) then $y$ changes as well. Therefore, when I calculate derivatives, then naturally I take $\frac{\partial x}{\partial y}$, because that is what I observe as a derivative at the level of the physical system (e.g. lowering current, rises the voltage). This means, that in the Jacobian and Hessian I expect to see $\frac{\partial x}{\partial y}$, rather than $\frac{\partial f(x,y(x))}{\partial y}$.