Consider $F: D \subset \mathbb{R}^2 \to \mathbb{R}$. If I call
$$ F(x,y) = 0$$
an implicit equation, then presumably, because I'm calling this an implicit equation, I'm implying there exists at least one function $g(x)$ over some interval $I$ which satisfies $F(x,g(x)) = 0$.
Let's consider
$$ x^2 + y^2 + 1 = 0 \tag{1}$$ This equation does not define an implicit function $y = g(x)$. Wikipedia would still call $(1)$ an implicit equation. I'd like to reserve the word for cases which actually do imply an implicit function. Anyways, why doesn't the implicit function theorem work on equation $(1)$? The wikipedia article is a little abstract and general for me, so I looked up an implicit function theorem in 2 variables.
The domain of $F$ is $\mathbb{R}^2$. The partial derivatives $$ F_x = 2x \\ F_y = 2y$$ exist and are continuous. $F_y = 0 $ only when $y = 0$. So it seems like an implicit function should exist according to the theorem (possibly on each side of $y = 0$ in the plane), even though we know one doesn't?
Why does
$$x^2 + y^2 - 1 = 0$$
satisfy the conditions of the theorem but $(1)$ doesn't? I must be missing something simple here
To apply the implicit function theorem, you have to fix a point $(a, b) \in \mathbb{R^2}$, such that $f(a, b) = 0$. But there is no such point in $\mathbb{R}^2$: Assume the validity of (1) for some $x, y \in \mathbb{R}^2$. $(1) \implies x^2 + y^2 = -1$. But if both $x$ and $y$ are in $\mathbb{R}$, their squares are nonnegative and likewise the sum of their squares ($x^2 + y^2$) too. Contradiction.
(Probably, there are such points in $\mathbb{C}^2 \setminus \mathbb{R}^2$, but these are not relevant.)
$x^2 + y^2 - 1 = 0$ defines the unit circle in the plane $\mathbb{R}^2$. Especially, there are points fulfilling $f(a, b) = 0$ (e. g. $a = 0$, $b = 1$) and with an invertible Jacobian matrix at this point (c. f. here)