Suppose $g(x)=f(x)m(x)^{-1}$ where $f(x)\in F[x]$ and $F$ has characteristic $0$ and $m(x)=(f,f')$. Then show that all roots of $g(x)$ are simple.
I assume g(x) has multiple roots that is $g(x)=(x-s)^nh(x)$ and after computing the derivative I concluded $(x-s)|g'(x)$. Then I consider $g(x)=fm^{-1}$ and use the fact that $f(x)=m(x)a(x)$ and $f'(x)=m(x)b(x)$. I rewrite $g'(x)=f(x)m(x)^{-2}+f'(x)m(x)^{-1}$. From this I want to conclude that the right hand side is not divided by $(x-s)$ but I am stuck at this part.
We'll assume for simplicity that $f$ is monic. Factor $f(x)$ in an algebraic closure as $(x-r_1)^{n_1}\cdots(x-r_k)^{n_k}$, where the $r_i$'s are distinct roots. You can check that $f'$ is divisible by $(x-r_1)^{n_1-1}\cdots(x-r_k)^{n_k-1}$. This product should be $m(x)$, and since $F$ has characteristic $0$, $m(x)\neq 0$ (assuming $f$ is nonconstant). But proving it's exactly equal to $n(x)$ is unnecessary; we just need that $m\mid f$ and $m$ is a multiple of the product above. Then, $f/m$ will be a factor of $(x-r_1)\cdots (x-r_k)$, which has every root with multiplicity $1$.