Scketch of an elementary proof. Assume $F\in C^2(\mathbb{R}^2, \mathbb{R})$ and
(changing sign to $F$ if needed)
$$F(0,0)=F_x(0,0)=F_y(0,0)=F_{xy}(0,0)=0\, ,$$
$$F_{xx}(0,0) <0\, ,\qquad F_{yy}(0,0) > 0\, .$$

Therefore there exist $\delta > 0$ and $\epsilon > 0$ such that
$F_{yy}(x,y) > 0$ for $|x| \le \eta$ and $|y|\le\epsilon$. So for all $|x|\le \eta$
the function $y\mapsto F(x,y)$ is strictly convex on the interval
$[-\epsilon,\epsilon]$. In particular $F(0,\pm\epsilon) >0$ because
$F(0,0)=F_y(0,0)=0$. Since $F(0,\pm\epsilon) >0$ and $F_{xx}(0,0) < 0$, we also
have by continuity $F(x,\pm\epsilon) >0$ and $F_{xx}(x,0) < 0$, for all
$|x|\le\delta$ for some $0 < \delta\le\eta$; thus $F(x,0) < 0$ for $0 < |x|\le
\delta$. Now, since for all $|x|\le\delta$ the function $y\mapsto F(x,y)$ is
strictly convex on the interval $[-\epsilon,\epsilon]$, positive at $y=\pm\epsilon$
and negative at $y=0$, for any $0 < |x|\le\delta$ we have $F(x,y)=0$ exactly for one
$0 < y < \epsilon$ and one $-\epsilon < y < 0$, always with $F _ y (x,y)\neq0$,
while $F(0,y)=0$ exactly for $y=0$ if $|y|\le\epsilon$. This proves that the trace
of the zero-set of $F$ on $[-\delta,\delta]\times [-\epsilon,\epsilon]$ is the union
of the graphs of two functions, $y_+: [-\delta,\delta]\to [-\epsilon,\epsilon]$ and
$y_-: [-\delta,\delta]\to [-\epsilon,\epsilon]$ defined so that $\operatorname{sgn}
y _ + (x)=\operatorname{sgn} x$ and $\operatorname{sgn}y _ - (x)=-\operatorname{sgn}
x$. Note that the fact that $\epsilon$ is arbitrary immediately implies that $y_+$
and $y_-$ are continuous at $x=0$ and vanish there. Actually, if we locate the zero-set of $F$ with a bit more care we also have that $y _ \pm (x) $
is derivable at $x=0$: this follows from the fact that $F$ satisfies an inequality locally at the origin:
$$\big(F_{xx}(0,0)+o(1) \big)x^2/2 +\big( F_{yy}(0,0)+o(1) \big)y^2/2 \le F(x,y) $$

$$\le \big(F _ {xx} (0,0) +o(1) \big) x^2/2 + \big( F_{yy} (0,0) +o(1) \big)y^2/2$$

so that $(x,y _ \pm (x))$ belongs to a very thin cone around the line $y= \pm F_{xx}(0,0)/F_{yy}(0,0)x$, meaning that $y _ \pm (x)$ is derivable at $x=0$ and $ y ' _ \pm (0)= -\sqrt {- \frac { F _ {xx}(0,0) } { F_ {yy} (0,0) } }\, . $
Moreover, since for all
$ (x,y) \in \{ F = 0 \} \cap [ -\delta, \delta] \times [-\epsilon,\epsilon]
\setminus\{ (0,0)\}$ we have $F _ y(x,y)\neq 0$ the standard Implicit Function
Theorem ensures that $y_+$ and $y_-$ are $C^0([ -\delta, \delta])\cap C^1([ -\delta, \delta]\setminus \{ 0
\}) $, with $$F_x(x,y_\pm(x))+F_y(x,y_\pm(x))\dot y_\pm(x)=0\quad , \quad \forall x\neq0 .$$ To prove that they are $C^1([ -\delta, \delta])$ note that for $|x|+|y|\to 0$
$$F_x(x,y)=F_{xx}(0,0)x+o(|x|+|y|)$$
$$F_y(x,y)=F_{yy}(0,0)y+o(|x|+|y|)\, .$$
Therefore we have
$$F _ {xx}(0,0)+y' _ \pm (x)\frac{y _ \pm(x)}{x}F _ {yy}(0,0)=\Big(1+|y' _ \pm(x)|\Big)\Big(1+\big|\frac{y _ \pm(x)}{x}\big|\Big)o(1)\, .$$
Since $\frac{y_\pm(x)}{x}=y_\pm'(0)+o(1)$, and $\pm y'_ \pm(x)\ge0$ this implies that

$y^\prime _ \pm (x) \to y^\prime _ \pm (0) $ for $ x \to 0 $ , so $ y^\prime _ \pm $ is continuous and $ y _ \pm \in C^1 ([ - \delta , \delta ] ) $ .