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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

$y'y^' _ \pm(x)\to pm (x) \to y' _ \pm(0)$ pm (0) $for$ x\to0$, x \to 0$ , so $y'y' _ \pm$ pm $is continuous and$ y _ \pm\in C^1(pm \in C^1 ([ - \delta, \delta])$. delta , \delta ] )$ .

3 added 76 characters in body; deleted 63 characters in body; added 17 characters in body

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

$$\le \le big(F _ {xx} (F_{xx}(0,0)+o(1) )0,0) +o(1) \big) x^2/2 + \big( F_{yy} (F_{yy}(0,0)+o(1) )y^2/2$$ 0,0) +o(1) \big)y^2/2$$so that (x,y _ \pm (x)) belongs to a very thin cone around the line y= \pm F_{xx}(0,0)/F_{yy}(0,0)x, so meaning that y _ \pm (x) is derivable at x=0 and and  y'y ' _ \pm(0)= pm (0)= -\sqrt{-\frac{F\sqrt {xx}(0,0)}{F_{yy}(0,0)}} - \frac { F _ {xx}(0,0) } { F_ {yy} (0,0) } }\ .  Moreover, since for all  (x,y) \in \{ F = 0 \} \cap [ -\delta, \delta] \times [-\epsilon,\epsilon] \setminus\{ (0,0)\} we have F _ y(x,y)\neq0 y(x,y)\neq 0 the standard Implicit Function Theorem ensures that y_+ and y_- are C^0([ -\delta, \delta])\cap C^1([ -\delta, \delta]\setminus \{ 0 \}) , with$$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 $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)$$x)\frac{y _ \pm(x)}{x}F _ {yy}(0,0)=\Big(1+|y' _ \pm(x)|\Big)\Big(1+\big|\frac{y _ \pm(x)}{x}\big|\Big)o(1)\ .$$ Since$\frac{y\pm(x)}{x}=y\pm'(0)+o(1)$, \frac{y_\pm(x)}{x}=y_\pm'(0)+o(1)$, and $\pm y'\pm(x)\ge0$ y'_ \pm(x)\ge0$this implies that$y'\pm(x)\to y'\pm(0)$y'\pm(x)\to y'\pm(0)$ for $x\to0$, so $y'\pm$ y'\pm$is continuous and$y_\pm\in y\pm\in C^1([ -\delta, \delta])$. rmk. More generally, I think$y_\pm$is$C^{k-1}$if$F$is$C^k$. 2 added 308 characters in body and$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:$$(F_{xx}(0,0)+o(1) )x^2/2 +( F_{yy}(0,0)+o(1) )y^2/2 \le F(x,y) \le (F_{xx}(0,0)+o(1) )x^2/2 + ( F_{yy}(0,0)+o(1) )y^2/2$$so that$(x,y _ \pm (x))$belongs to a very thin cone around the line$y= \pm F_{xx}(0,0)/F_{yy}(0,0)x$, so$y _ \pm (x)$is derivable at$x=0$and and$y'\pm(0)= -\sqrt{-\frac{F{xx}(0,0)}{F_{yy}(0,0)}} }\ .$, both for$y(x)=y_+(x)$and for$y(x)=y_-(x)$$F_{xx}(0,0)+y'\pm (x)\frac{y\pm(x)}{x}F_{yy}(0,0)=\Big(1+|y'\pm(x)|\Big)\Big(1+\big|\frac{y\pm(x)}{x}\big|\Big)o(1)$$Since $\frac{y(x)}{x}=y'(\xi)$ for some $| \xi | < |x| \frac{y\pm(x)}{x}=y\pm'(0)+o(1)$, and $, \pm y'\pm(x)\ge0$ this implies that $|y'(x)|$ has a limit y'\pm(x)\to y'\pm(0)$for$x\to0$; and since$\pm y'_ \pm(x)\ge0$, one conclude that there exist the limits$$\lim _ {x\to0}\ y'_ \pm(x)=\pm\sqrt{ -\frac{F_{xx}(0,0)}{F_{yy}(0,0)} }\ ,$$which implies that x\to0$, so $y_\pm$ y'\pm$is derivable at$x=0$too, with the same value, proving that continuous and$y_\pm\in C^1([ -\delta, \delta])\$.

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