$\newcommand\R{\mathbb R}\newcommand\N{\mathbb N}$Yes, for any real $a$ and any $k\in\{0,1,\dots\}$, we have $$\text{$f^{(k)}(a)=0$ if $h$ does not have all the derivatives at $a$.}\tag{1}$$
Indeed, take any $a\in\R$ and suppose that there is a sequence $(x_n)_{n\in\N}$ in $\R$ converging to $a$ such that for all $n$ we have $x_n\ne a$ and $f(x_n)=0$. This sequence may be assumed to be strictly monotone. So, by Rolle's theorem, for each $k\in\{0,1,\dots\}$ there is a sequence $(x_{k;n})_{n\in\N}$ such that $x_{k;n}\to a$ (as $n\to\infty$) and $f^{(k)}(x_{k;n})=0$ for all $n\in\N$, so that $f^{(k)}(a)=0$.
So, if $f^{(k)}(a)\ne0$, then for all $x$ in a neighborhood $V$ of $a$ we have $f(x)\ne0$; hence, $h=(hf)/f$ is smooth on $V$ and therefore has all the derivatives at $a$. So, (1) is proved.