We have a simple generalization of the lemniscate of Bernoulli defined by a function $f : \mathbb{R}^2 \to \mathbb{R}$: $$ f(x, y) = (x^2 + y^2)^2 - 2(x^2 - y^2) $$ The level curve of this function when $f(x, y) = 0$ gives the lemniscate, but for $f(x, y) = C$ we get different shapes.
I would like to find the set of points at which the implicit function theorem does not apply to the equation $f(x,y) = C$ to yield $y = y(x)$.
In order to apply the theorem, the function we're examining has to be in the form $g(x, y) = 0$, so define $g(x, y) = f(x, y) - C$. But then $g(0, 0) = -C \neq 0$ unless $C = 0$, so we cannot apply the theorem for values of $C$ other than $0$... This feels like an incorrect approach, since I feel like the theorem is applicable. (Especially because it's possible to solve explicitly for $y$ in terms of $x$, but it takes a bit of work.)
On the other hand, if we just ignore that $f$ is not in the right form, and go on with the theorem, we get that $f(0, 0) = 0$, so on that front we're OK. Next we have to look at the partial derivative with respect to $y$. $$ \partial_y f(x, y) = 4y(x^2 + y^2 + 1) $$ If we consider $\partial_y f(x, y) = 0$, then either $y = 0$ or $x^2 + y^2 + 1 = 0$. The latter case is a contradiction, because $x^2 \geq 0$ and $y^2 \geq 0$, so we cannot choose $x, y$ to satisfy the equation. So we can apply the implicit function theorem for all points $(x, y)$ provided that $y \neq 0$.
Is this a fair assessment? Does it make sense to ignore that we don't have the function in the correct form?
You are correct: if you want to use IFT to express $y$ as a function of $x$ near $(x_0, y_0)$ on the curve $f(x,y) = C$ (where $f$ is continuously differentiable), what you need is $\partial f/\partial y (x_0, y_0)\ne 0$. The fact that the equation is not in the form $f(x,y) = 0$ with $f(0,0)=0$ is not relevant, as you can write it as $f(s+x_0,t+y_0) - C = 0$.