$\textbf{Question : }$Find all $f\in C^{\infty}(\mathbb{R})$ such that, as a distribution, $f\cdot\delta'=0.$
$\textbf{My Attempt : }$
Since $f\cdot\delta'=0$, for every $\phi\in\mathscr{D}(\mathbb{R})$, we have $0=\langle f\cdot\delta',\phi\rangle=\langle \delta',f\cdot\phi \rangle=-\langle \delta,(f\cdot\phi)' \rangle.$
Therefore, $(f\cdot\phi)'(0)=f(0)\phi'(0)+f'(0)\phi(0)=0$ for all $\phi\in\mathscr{D}(\mathbb{R})$.
Hence, the solution set is $\{f\in C^{\infty}(\mathbb{R}):f(0)=f'(0)=0\}.$
Are the above arguments correct ? If yes then can we find somewhat more explicit solution set ?
Another way: $$ 0 = f\delta' = (f\delta)' - f'\delta = (f(0)\delta)' - f'(0)\delta = f(0)\delta' - f'(0)\delta. $$ Since $\delta$ and $\delta'$ are linearly independent, we must have $f(0)=f'(0)=0.$