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$\newcommand\R{\mathbb R}\newcommand\N{\mathbb N}$Here are answers to your three questions (the latter two of them partial).

Answer 1: Yes, for any real $a$ and any $k\in\{0,1,\dots\}$, \begin{equation*} \text{if $h$ does not have all the derivatives at $a$, then $f^{(k)}(a)=0$.}\tag{1} \end{equation*}

Indeed, say that $h$ is bad at $a$ if $h$ does not have all the derivatives at $a$.

Take any $a\in\R$. Suppose that $h$ is bad at $a$. Take then the smallest $k\in\{0,1,\dots\}$ such that $f^{(k)}(a)\ne0$, if such a $k$ exists. Let \begin{equation*} g:=fh \end{equation*} and \begin{equation*} F(x):=\frac{f(x)}{(x-a)^k}, \quad G(x):=\frac{g(x)}{(x-a)^k} \end{equation*} for real $x\ne a$, with $F(a):=f^{(k)}(a)/k!$ and $G(a):=g^{(k)}(a)/k!$. By a Taylor formula, if $k\ge1$, then for all real $x$ \begin{equation*} F(x)=\frac1{(k-1)!}\int_0^1(1-s)^{k-1}f^{(k)}(a+(x-a)s)\,ds \end{equation*} and hence for all nonnegative integers $l$ \begin{equation*} F^{(l)}(x)=\frac1{(k-1)!}\int_0^1(1-s)^{k-1}s^lf^{(k+l)}(a+(x-a)s)\,ds, \end{equation*} with the similar formulas for $G(x)$ and $G^{(l)}(x)$. So, if $k\ge1$, then $F$ and $G$ are smooth, $F(a)\ne0$, and hence $F\ne0$ and $h=G/F$ on a neighborhood $V$ of $a$ (the equality $h(a)=G(a)/F(a)$ follows by continuity); the same conclusions obviously hold for $k=0$. So, $h$ is smooth on $V$, which contradicts the assumption that $h$ is bad at $a$. So, as claimed, there is no $k\in\{0,1,\dots\}$ such that $f^{(k)}(a)\ne0$.

Answer 2 (partial): Now it follows that $g_k:=f^{(k)}h$ must be differentiable. Indeed, take any real $a$. If $h$ has all the derivatives at $a$, then so does $g_k$. If $h$ does not have all the derivatives at $a$, then, by (1), $f'(x)=o(|x-a|)$ as $x\to a$, so that $g_k(x)=o(|x-a|)$ as $x\to a$, so that $g_k'(a)=0$.

Answer 3 (partial): Finally, let $g_{k,l}:=f^{(k)}h^{(l)}$ wherever $h^{(l)}$ exists, with $g_{k,l}:=0$ elsewhere. Take any real $a$ such that $h$ does not have all the derivatives at $a$. Note that $h'$ is bounded on the set where $h'$ exists, since $h$ is Lipschitz. By (1), $f^{(k)}(x)=o(|x-a|)$ as $x\to a$, so that $g_{k,1}(x)=o(|x-a|)$ as $x\to a$, and hence $g_{k,1}'(a)=0$. So, $g_{k,1}$ is differentiable.

$\newcommand\R{\mathbb R}\newcommand\N{\mathbb N}$Here are answers to your three questions (the latter two of them partial).

Answer 1: Yes, for any real $a$ and any $k\in\{0,1,\dots\}$, \begin{equation*} \text{if $h$ does not have all the derivatives at $a$, then $f^{(k)}(a)=0$.}\tag{1} \end{equation*}

Indeed, say that $h$ is bad at $a$ (or, equivalently, that $a$ is a bad point of $h$0 if $h$ does not have all the derivatives at $a$.

Take any $a\in\R$. Suppose that $h$ is bad at $a$. Take then the smallest $k\in\{0,1,\dots\}$ such that $f^{(k)}(a)\ne0$, if such a $k$ exists. Let \begin{equation*} g:=fh \end{equation*} and \begin{equation*} F(x):=\frac{f(x)}{(x-a)^k}, \quad G(x):=\frac{g(x)}{(x-a)^k} \end{equation*} for real $x\ne a$, with $F(a):=f^{(k)}(a)/k!$ and $G(a):=g^{(k)}(a)/k!$. By a Taylor formula, if $k\ge1$, then for all real $x$ \begin{equation*} F(x)=\frac1{(k-1)!}\int_0^1(1-s)^{k-1}f^{(k)}(a+(x-a)s)\,ds \end{equation*} and hence for all nonnegative integers $l$ \begin{equation*} F^{(l)}(x)=\frac1{(k-1)!}\int_0^1(1-s)^{k-1}s^lf^{(k+l)}(a+(x-a)s)\,ds, \end{equation*} with the similar formulas for $G(x)$ and $G^{(l)}(x)$. So, if $k\ge1$, then $F$ and $G$ are smooth, $F(a)\ne0$, and hence $F\ne0$ and $h=G/F$ on a neighborhood $V$ of $a$ (the equality $h(a)=G(a)/F(a)$ follows by continuity); the same conclusions obviously hold for $k=0$. So, $h$ is smooth on $V$, which contradicts the assumption that $h$ is bad at $a$. So, as claimed, there is no $k\in\{0,1,\dots\}$ such that $f^{(k)}(a)\ne0$.

We have actually proved more than (1): \begin{equation*} \text{if $a$ is in the closure of the set of all bad points of $h$, } \\ \text{then $f^{(k)}(a)=0$ for all $k\in\{0,1,\dots\}$.} \end{equation*}

Answer 2 (partial): Now it follows that $g_k:=f^{(k)}h$ must be differentiable. Indeed, take any real $a$. If $h$ has all the derivatives at $a$, then so does $g_k$. If $h$ does not have all the derivatives at $a$, then, by (1), $f'(x)=o(|x-a|)$ as $x\to a$, so that $g_k(x)=o(|x-a|)$ as $x\to a$, so that $g_k'(a)=0$.

Answer 3 (partial): Finally, let $g_{k,l}:=f^{(k)}h^{(l)}$ wherever $h^{(l)}$ exists, with $g_{k,l}:=0$ elsewhere. Take any real $a$ such that $h$ does not have all the derivatives at $a$. Note that $h'$ is bounded on the set where $h'$ exists, since $h$ is Lipschitz. By (1), $f^{(k)}(x)=o(|x-a|)$ as $x\to a$, so that $g_{k,1}(x)=o(|x-a|)$ as $x\to a$, and hence $g_{k,1}'(a)=0$. So, $g_{k,1}$ is differentiable.

$\newcommand\R{\mathbb R}\newcommand\N{\mathbb N}$Here are answers to your three questions (the latter two of them partial).

Answer 1: Yes, for any real $a$ and any $k\in\{0,1,\dots\}$, \begin{equation*} \text{if $h$ does not have all the derivatives at $a$, then $f^{(k)}(a)=0$.}\tag{1} \end{equation*}

Indeed, say that $h$ is bad at $a$ if $h$ does not have all the derivatives at $a$.

Take any $a\in\R$. Suppose that $h$ is bad at $a$. Take then the smallest $k\in\{0,1,\dots\}$ such that $f^{(k)}(a)\ne0$, if such a $k$ exists. Let \begin{equation*} g:=fh \end{equation*} and \begin{equation*} F(x):=\frac{f(x)}{(x-a)^k}, \quad G(x):=\frac{g(x)}{(x-a)^k} \end{equation*} for real $x\ne a$, with $F(a):=f^{(k)}(a)/k!$ and $G(a):=g^{(k)}(a)/k!$. Then, by a Taylor formula, for all real $x$ \begin{equation*} F(x)=\frac1{(k-1)!}\int_0^1(1-s)^{k-1}f^{(k)}(a+(x-a)s)\,ds \end{equation*} if $k\ge1$, with the similar formula for $G(x)$. So, if $k\ge1$, then $F$ and $G$ are smooth, $F(a)\ne0$, and hence $F\ne0$ and $h=G/F$ on a neighborhood $V$ of $a$ (the equality $h(a)=G(a)/F(a)$ follows by continuity); the same conclusions obviously hold for $k=0$. So, $h$ is smooth on $V$, which contradicts the assumption that $h$ is bad at $a$. So, as claimed, there is no $k\in\{0,1,\dots\}$ such that $f^{(k)}(a)\ne0$.

Answer 2 (partial): Now it follows that $g_k:=f^{(k)}h$ must be differentiable. Indeed, take any real $a$. If $h$ has all the derivatives at $a$, then so does $g_k$. If $h$ does not have all the derivatives at $a$, then, by (1), $f'(x)=o(|x-a|)$ as $x\to a$, so that $g_k(x)=o(|x-a|)$ as $x\to a$, so that $g_k'(a)=0$.

Answer 3 (partial): Finally, let $g_{k,l}:=f^{(k)}h^{(l)}$ wherever $h^{(l)}$ exists, with $g_{k,l}:=0$ elsewhere. Take any real $a$ such that $h$ does not have all the derivatives at $a$. Note that $h'$ is bounded on the set where $h'$ exists, since $h$ is Lipschitz. By (1), $f^{(k)}(x)=o(|x-a|)$ as $x\to a$, so that $g_{k,1}(x)=o(|x-a|)$ as $x\to a$, and hence $g_{k,1}'(a)=0$. So, $g_{k,1}$ is differentiable.

$\newcommand\R{\mathbb R}\newcommand\N{\mathbb N}$Here are answers to your three questions (the latter two of them partial).

Answer 1: Yes, for any real $a$ and any $k\in\{0,1,\dots\}$, \begin{equation*} \text{if $h$ does not have all the derivatives at $a$, then $f^{(k)}(a)=0$.}\tag{1} \end{equation*}

Indeed, say that $h$ is bad at $a$ if $h$ does not have all the derivatives at $a$.

Take any $a\in\R$. Suppose that $h$ is bad at $a$. Take then the smallest $k\in\{0,1,\dots\}$ such that $f^{(k)}(a)\ne0$, if such a $k$ exists. Let \begin{equation*} g:=fh \end{equation*} and \begin{equation*} F(x):=\frac{f(x)}{(x-a)^k}, \quad G(x):=\frac{g(x)}{(x-a)^k} \end{equation*} for real $x\ne a$, with $F(a):=f^{(k)}(a)/k!$ and $G(a):=g^{(k)}(a)/k!$. By a Taylor formula, if $k\ge1$, then for all real $x$ \begin{equation*} F(x)=\frac1{(k-1)!}\int_0^1(1-s)^{k-1}f^{(k)}(a+(x-a)s)\,ds \end{equation*} and hence for all nonnegative integers $l$ \begin{equation*} F^{(l)}(x)=\frac1{(k-1)!}\int_0^1(1-s)^{k-1}s^lf^{(k+l)}(a+(x-a)s)\,ds, \end{equation*} with the similar formulas for $G(x)$ and $G^{(l)}(x)$. So, if $k\ge1$, then $F$ and $G$ are smooth, $F(a)\ne0$, and hence $F\ne0$ and $h=G/F$ on a neighborhood $V$ of $a$ (the equality $h(a)=G(a)/F(a)$ follows by continuity); the same conclusions obviously hold for $k=0$. So, $h$ is smooth on $V$, which contradicts the assumption that $h$ is bad at $a$. So, as claimed, there is no $k\in\{0,1,\dots\}$ such that $f^{(k)}(a)\ne0$.

Answer 2 (partial): Now it follows that $g_k:=f^{(k)}h$ must be differentiable. Indeed, take any real $a$. If $h$ has all the derivatives at $a$, then so does $g_k$. If $h$ does not have all the derivatives at $a$, then, by (1), $f'(x)=o(|x-a|)$ as $x\to a$, so that $g_k(x)=o(|x-a|)$ as $x\to a$, so that $g_k'(a)=0$.

Answer 3 (partial): Finally, let $g_{k,l}:=f^{(k)}h^{(l)}$ wherever $h^{(l)}$ exists, with $g_{k,l}:=0$ elsewhere. Take any real $a$ such that $h$ does not have all the derivatives at $a$. Note that $h'$ is bounded on the set where $h'$ exists, since $h$ is Lipschitz. By (1), $f^{(k)}(x)=o(|x-a|)$ as $x\to a$, so that $g_{k,1}(x)=o(|x-a|)$ as $x\to a$, and hence $g_{k,1}'(a)=0$. So, $g_{k,1}$ is differentiable.

$\newcommand\R{\mathbb R}\newcommand\N{\mathbb N}$Here are answers to your three questions (the latter two of them partial).

Answer 1: Yes, for any real $a$ and any $k\in\{0,1,\dots\}$, \begin{equation*} \text{if $h$ does not have all the derivatives at $a$, then $f^{(k)}(a)=0$.}\tag{1} \end{equation*}

Indeed, say that $h$ is bad at $a$ if $h$ does not have all the derivatives at $a$.

Take any $a\in\R$. Suppose that $h$ is bad at $a$. Take then the smallest $k\in\{0,1,\dots\}$ such that $f^{(k)}(a)\ne0$, if such a $k$ exists. Let \begin{equation*} g:=fh \end{equation*} and \begin{equation*} F(x):=\frac{f(x)}{(x-a)^k}, \quad G(x):=\frac{g(x)}{(x-a)^k} \end{equation*} for real $x\ne a$, with $F(a):=f^{(k)}(a)/k!$ and $G(a):=g^{(k)}(a)/k!$. Then, by a Taylor formula, for all real $x$ \begin{equation*} F(x)=\frac1{(k-1)!}\int_0^1(1-s)^{k-1}f^{(k)}(a+(x-a)s)\,ds \end{equation*} if $k\ge1$, with the similar formula for $G(x)$. So, if $k\ge1$, then $F$ and $G$ are smooth, $F(a)\ne0$, and hence $F\ne0$ and $h=G/F$ on a neighborhood $V$ of $a$ (the equality $h(a)=G(a)/F(a)$ follows by continuity); the same conclusions obviously hold for $k=0$. So, $h$ is smooth on $V$, which contradicts the assumption that $h$ is bad at $a$. So, as claimed, there is no $k\in\{0,1,\dots\}$ such that $f^{(k)}(a)\ne0$.

Answer 2 (partial): Now it follows that $g_k:=f^{(k)}h$ must be differentiable. Indeed, take any real $a$. If $h$ has all the derivatives at $a$, then so does $g_k$. If $h$ does not have all the derivatives at $a$, then, by (1), $f'(x)=o(|x-a|)$ as $x\to a$, so that $g_k(x)=o(|x-a|)$ as $x\to a$, so that $g_k'(a)=0$.

Answer 3 (partial): Finally, let $g_{k,l}:=f^{(k)}h^{(l)}$ wherever $h^{(l)}$ exists, with $g_{k,l}:=0$ elsewhere. Take any real $a$ such that $h$ does not have all the derivatives at $a$. Note that $h'$ is bounded on the set where $h'$ exists, since $h$ is Lipschitz. By (1), $f^{(k)}(x)=o(|x-a|)$ as $x\to a$, so that $g_{k,1}(x)=o(|x-a|)$ as $x\to a$, and hence $g_{k,1}'(a)=0$. So, $g_{k,1}$ is differentiable.

However, $g_{0,2}$ is not continuous in general. E.g., let $f(x):=e^{-1/|x|}$ for $x\ne0$, with $f(0):=0$. Let $$\tilde h(x):=1(x>0)\int_0^x h_1(u)\,du,$$ where $$h_1(u):=\sum_{n\in\N}(-1)^n\,1(\tfrac1{n+1}<u<\tfrac1n).$$ Then $h_1$ is Lipschitz and $|\tilde h''(\tfrac1n)|=\infty$ for all $n\in\N$. Approximate $h_1$ closely enough by a Lipschitz function $h$ smooth enough on $(0,\infty)$ so that $|h''(\tfrac1n)|>e^{2n}$ for all $n$. Then $|g_{0,2}(\tfrac1n)|>e^{-n}e^{2n}\to\infty$ as $n\to\infty$. So, as claimed, $g_{0,2}$ is not continuous.

$\newcommand\R{\mathbb R}\newcommand\N{\mathbb N}$Here are answers to your three questions (the latter two of them partial).

Answer 1: Yes, for any real $a$ and any $k\in\{0,1,\dots\}$, \begin{equation*} \text{if $h$ does not have all the derivatives at $a$, then $f^{(k)}(a)=0$.}\tag{1} \end{equation*}

Indeed, say that $h$ is bad at $a$ if $h$ does not have all the derivatives at $a$.

Take any $a\in\R$. Suppose that $h$ is bad at $a$. Take then the smallest $k\in\{0,1,\dots\}$ such that $f^{(k)}(a)\ne0$, if such a $k$ exists. Let \begin{equation*} g:=fh \end{equation*} and \begin{equation*} F(x):=\frac{f(x)}{(x-a)^k}, \quad G(x):=\frac{g(x)}{(x-a)^k} \end{equation*} for real $x\ne a$, with $F(a):=f^{(k)}(a)/k!$ and $G(a):=g^{(k)}(a)/k!$. Then, by a Taylor formula, for all real $x$ \begin{equation*} F(x)=\frac1{(k-1)!}\int_0^1(1-s)^{k-1}f^{(k)}(a+(x-a)s)\,ds \end{equation*} if $k\ge1$, with the similar formula for $G(x)$. So, if $k\ge1$, then $F$ and $G$ are smooth, $F(a)\ne0$, and hence $F\ne0$ and $h=G/F$ on a neighborhood $V$ of $a$ (the equality $h(a)=G(a)/F(a)$ follows by continuity); the same conclusions obviously hold for $k=0$. So, $h$ is smooth on $V$, which contradicts the assumption that $h$ is bad at $a$. So, as claimed, there is no $k\in\{0,1,\dots\}$ such that $f^{(k)}(a)\ne0$.

Answer 2 (partial): Now it follows that $g_k:=f^{(k)}h$ must be differentiable. Indeed, take any real $a$. If $h$ has all the derivatives at $a$, then so does $g_k$. If $h$ does not have all the derivatives at $a$, then, by (1), $f'(x)=o(|x-a|)$ as $x\to a$, so that $g_k(x)=o(|x-a|)$ as $x\to a$, so that $g_k'(a)=0$.

Answer 3 (partial): Finally, let $g_{k,l}:=f^{(k)}h^{(l)}$ wherever $h^{(l)}$ exists, with $g_{k,l}:=0$ elsewhere. Take any real $a$ such that $h$ does not have all the derivatives at $a$. Note that $h'$ is bounded on the set where $h'$ exists, since $h$ is Lipschitz. By (1), $f^{(k)}(x)=o(|x-a|)$ as $x\to a$, so that $g_{k,1}(x)=o(|x-a|)$ as $x\to a$, and hence $g_{k,1}'(a)=0$. So, $g_{k,1}$ is differentiable.