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Let $f$ be a $C^2$ function defined on a neighborhood of $0$ in $\mathbb R^n$ such that $f(0)=0, df(0)\not=0$. By the Implicit Function Theorem, it is easy to get (after a rotation) that near 0 $$ f(x)=\bigl(x_n-\alpha(x')\bigr) e(x),\quad\text{$\alpha\in C^2$ near the origin in $\mathbb R^{n-1}$, $e\in C^1$.} $$ Moreover I claim that for $T=(t',t_n)\in \mathbb R^{n-1}\times \mathbb R$ $$ f''(x',\alpha(x'))T^2=-e(x', \alpha(x'))\alpha''(x')(t')^2 +2 (e'(x)\cdot T)(t_n-\alpha'(x')\cdot t'), $$ where $g''$ stands for the symmetric matrix of second derivatives. It would be a simple consequence of Leibniz formula if $e$ were $C^2$, but $e$ is no better than $C^1$. I guess a "direct" proof would do.

Let $f$ be a $C^2$ function defined on a neighborhood of $0$ in $\mathbb R^n$ such that $f(0)=0, df(0)\not=0$. By the Implicit Function Theorem, it is easy to get (after a rotation) that near 0 $$ f(x)=\bigl(x_n-\alpha(x')\bigr) e(x),\quad\text{$\alpha\in C^2$ near the origin in $\mathbb R^{n-1}$, $e\in C^1$.} $$ Moreover I claim that for $T=(t',t_n)\in \mathbb R^{n-1}\times \mathbb R$ $$ f''(x',\alpha(x'))T^2=-e(x', \alpha(x'))\alpha''(x')(t')^2 +2 (e'(x', \alpha(x'))\cdot T)(t_n-\alpha'(x')\cdot t'), $$ where $g''$ stands for the symmetric matrix of second derivatives. It would be a simple consequence of Leibniz formula if $e$ were $C^2$, but $e$ is no better than $C^1$. I guess a "direct" proof would do.

Let $f$ be a $C^2$ function defined on a neighborhood of $0$ in $\mathbb R^n$ such that $f(0)=0, df(0)\not=0$. By the Implicit Function Theorem, it is easy to get (after a rotation) that near 0 $$ f(x)=\bigl(x_n-\alpha(x')\bigr) e(x),\quad\text{$\alpha\in C^2$ near the origin in $\mathbb R^{n-1}$, $e\in C^1$.} $$ Moreover I claim that $$ f''(x',\alpha(x'))=-e(x', \alpha(x'))\alpha''(x'), $$ where $f''$ stands for the symmetric matrix of second derivatives. It would be a simple consequence of Leibniz formula if $e$ were $C^2$, but $e$ is no better than $C^1$. I guess a "direct" proof would do.

Let $f$ be a $C^2$ function defined on a neighborhood of $0$ in $\mathbb R^n$ such that $f(0)=0, df(0)\not=0$. By the Implicit Function Theorem, it is easy to get (after a rotation) that near 0 $$ f(x)=\bigl(x_n-\alpha(x')\bigr) e(x),\quad\text{$\alpha\in C^2$ near the origin in $\mathbb R^{n-1}$, $e\in C^1$.} $$ Moreover I claim that for $T=(t',t_n)\in \mathbb R^{n-1}\times \mathbb R$ $$ f''(x',\alpha(x'))T^2=-e(x', \alpha(x'))\alpha''(x')(t')^2 +2 (e'(x)\cdot T)(t_n-\alpha'(x')\cdot t'), $$ where $g''$ stands for the symmetric matrix of second derivatives. It would be a simple consequence of Leibniz formula if $e$ were $C^2$, but $e$ is no better than $C^1$. I guess a "direct" proof would do.

Let $f$ be a $C^2$ function defined on a neighborhood of $0$ in $\mathbb R^n$ such that $f(0)=0, df(0)\not=0$. By the Implicit Function Theorem, it is easy to get (after a rotation) that near 0 $$ f(x)=\bigl(x_n-\alpha(x')\bigr) e(x),\quad\text{$\alpha\in C^2$ near the origin in $\mathbb R^{n-1}$, $e\in C^1$.} $$ Moreover I claim that $$ f''(x',\alpha(x'))=-e(x', \alpha(x'))\alpha''(x'). $$ It would be a simple consequence of Leibniz formula if $e$ were $C^2$, but $e$ is no better than $C^1$. I guess a "direct" proof would do.

Let $f$ be a $C^2$ function defined on a neighborhood of $0$ in $\mathbb R^n$ such that $f(0)=0, df(0)\not=0$. By the Implicit Function Theorem, it is easy to get (after a rotation) that near 0 $$ f(x)=\bigl(x_n-\alpha(x')\bigr) e(x),\quad\text{$\alpha\in C^2$ near the origin in $\mathbb R^{n-1}$, $e\in C^1$.} $$ Moreover I claim that $$ f''(x',\alpha(x'))=-e(x', \alpha(x'))\alpha''(x'), $$ where $f''$ stands for the symmetric matrix of second derivatives. It would be a simple consequence of Leibniz formula if $e$ were $C^2$, but $e$ is no better than $C^1$. I guess a "direct" proof would do.