Proving the restriction of a curve to a certain set of points is a regular curve Let $F: \mathbb{R^2}\mapsto \mathbb{R}$ be a differentiable function. Consider $(x_0, y_0) \in \mathbb{R}^2$ such that $F(x_0, y_0)=0$ and $F_x^2(x_0, y_0) +F_y^2(x_0, y_0) \neq 0$. Prove that the set of points $(x, y)$ from $\mathbb{R}^2$ close to $(x_0, y_0)$ such that $F(x, y) = 0$ is the trace of a regular curve. 
I'm having trouble seeing where to start. (also, here, a parametrized differentiable curve $a(t)$ is said to be regular if for all t in it's domain, $a'(t) \neq 0 $).
 A: I assume that $F(x, y) \in C^1(U, \Bbb R)$ for some open $U \subset \Bbb R$ with $(x_0, y_0) \in U$; then:
The condition
$F_x^2(x_0, y_0) + F_y^2(x_0, y_0) \ne 0 \tag 1$
implies that at least one of
$F_x(x_0, y_0) \ne 0, \tag 2$
$F_y(x_0, y_0) \ne 0 \tag 3$
binds.  We can then suppose without loss of generality that (3) holds; then it follows from the implicit function theorem that there exists an open interval $V \ni x_0$ in $\Bbb R$ and a function $y(x) \in C^1(V, \Bbb R)$  with $y(x_0) = y_0$ such that
$F(x, y(x)) = 0 \tag 4$
holds in $V$.  We may then define the $C^1$ curve
$a(x) = (x, y(x)), \tag 5$
where $x \in V$; we see that
$a'(x) = \begin{pmatrix} 1 \\ y'(x) \end{pmatrix} \ne 0; \tag 6$
identifying the parameter $t$ with $x$ tidies up the result and makes in conform notationally with the text of the question.
In the event that $F_y(x, y) = 0$ we can perform an argument parallel to that above with the roles of $F_x$ and $F_y$ reversed; in either event we find that, since $a(t) \subset \{(x, y) \mid F(x, y) = 0 \}$, the set $F(x, y) = 0$ is "the trace" of the curve $a(t)$.
Another, and I think more elegant, way to proceed is to consider the vector field
$X(x, y) = \begin{pmatrix} F_y(x, y) \\ -F_x(x, y) \end{pmatrix} \tag 7$
defined in a neighborhood $U$ of $(x_0, y_0)$; that $F(x, y) \in C^1(U, \Bbb R)$ implies $F$ is locally Lipschitz on $U$; see my answer to this question.  Then in some neighborhood $W \ni x_0$ of $\Bbb R$ there is a unique solution $a(t) = (x(t), y(t))$ to
$\begin{pmatrix} x' \\ y' \end{pmatrix} = X(x, y) = \begin{pmatrix} F_y(x, y) \\ -F_x(x, y) \end{pmatrix} \tag 8$
with $a(t_0) = (x_0, y_0)$.  Since
$\alpha'(t) \cdot (F_x, F_y)^T = F_y F_x - F_x F_y = 0 \tag 9$
and $\alpha(t_0) = (x_0, y_0)$, we see that $a(t)$ is contained in the level set $F(x, y) = 0$.  Since
$\alpha'(t) \cdot \alpha'(t) = F_x^2 + F_y^2 \ne 0, \tag{10}$
we see that
$\alpha'(t) \ne 0, \tag{11}$
so $\alpha(t)$ is a regular curve.  And again, the curve $F(x, y) = 0$ passing through $(x_0, y_0)$ is the trace of $a(t)$.
