$\newcommand\R{\mathbb R}\newcommand\N{\mathbb N}$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 l'Hospital's rule) $F$ 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). 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 $f^{(k)}(a)\ne0$.