I was given a theorem to solve this problem. The theorem states:
Assume that $F: \mathbb{R^2}\to\mathbb{R}$ is $C^1$, and let $S$ := {$x ∈ \mathbb{R^2} : F(x) = 0$}. If a ∈ $S$ and $∇F(a)≠0$, then there exists some $r > 0$ such that $B(r,a)∩S$ is a $C^1$ graph.
The solution shows that when $a$ is the point where the two branches on the graph intersects, $a$ must be a critical point of $F$, which is the part I don't understand. What's the significance of the intersected point, and how do we know that $a$ is singular, and not regular at that point?
Let $a$ be the intersection point, and suppose for the sake of contradiction that $\nabla F(a) \neq 0$. Then by your theorem, you can find an $r> 0$ such that $S \cap B(r,a)$ is the GRAPH of a $C^1$ function. This is absurd because near the point $a$, the graph would look like a "cross" $\times$, which means it fails the "vertical line test". Hence, $S \cap B(r,a)$ is NOT the graph of a $C^1$ function. Thus, we arrived at a contradiction. This means $\nabla F(a) = 0$, just as claimed in the notes.