Prove that curve defined by $F (x_0, y_0) = 0$ is regular

Question

The following exercise comes from Keti Tenenblat's introductory text on differential geometry. It asks:

Let $F: \mathbb {R}^2 \rightarrow \mathbb {R}$ be a differentiable function. Let $(x_0, y_0) \in \mathbb {R}$ be such that $F (x_0, y_0) = 0$ and $F_x^2 (x_0, y_0) + F_y^2 (x_0, y_0) \neq 0$. Show that the set of points $(x, y) \in \mathbb {R}^2$ which are close* to $(x_0, y_0)$ such that $F (x, y) = 0$ is the tracing of a regular curve.

Here is what I've tried: consider that the set of points which satisfy $F (x, y) = 0 $ is parameterized by a function $\alpha: I \rightarrow \mathbb {R}^2 $ such that $F (x, y) = F (x (t), y (t)) = F (\alpha (t)) = 0$. Taking the derivative of $F $ w.r.t. $t $, we get $$ \frac {dF}{dt} = \nabla F \bullet \alpha'(t). $$

Since $F $ is identically zero for all of the parametrized points, we have $$ \nabla F \bullet \alpha'(t) = 0 \Rightarrow x'(t) F_x = - y'(t) F_y . $$

Now, since we have assumed the gradient of $F $ to be nonzero, we have that this equation holds if $\nabla F $ is orthogonal to $\alpha'(t) $, or $\alpha'(t) $ is identically 0. But proving that it is not zero is exactly what I want, and I don't really see anything else to consider.

What would make me draw the conclusion that $\alpha'(t) $ is not zero? What have I missed?

Answer 1

Gradient $\nabla$ is produced with \nabla and not \delta.

It is a straightforward application of the implicit function theorem. Assume without loss of generality that $F_y(x_0,y_0) \neq 0$. Then the IFT gives intervals $I$ and $J$ around $x_0$ and $y_0$ such that for all $x \in I$ there is an unique $y(x) \in J$ with $F(x,y(x)) = 0$, and the map $y:I \to J$ so defined if of same differentiability class of $F$.

Clearly $\alpha\colon I \to \Bbb R^2$ given by $\alpha(x) = (x,y(x))$ is regular and does the job.