Let $f(x)$ be a monic irreducible polynomial over $\mathbb{Z}$ and let $\alpha$ be a root of $f$. Then I have to show that $f(r)~|~disc(\alpha)$, when $f'(x)$ has a root $r$ in $\mathbb{Z}$.
These are the things I have tried so far.
(1) First I tried to understand how one can understand $N(f'(\alpha))$.
(2) Writing $f'(x)=(x-r)g(x)$, with $g(x)\in\mathbb{Z}[x]$ and using the division algorithm one can write $f(x)=(x-r)^2q(x)+R$, for some $q(x)\in \mathbb{Z}[x]$ and $R\in\mathbb{Z}\setminus\{0\}$. Since $f(\alpha)=0$, it follows that $f(r)=R=-(\alpha-r)^2q(\alpha)$. This way we can get a relation between $g$ and $q$ and $q'$, i.e., $g=2q+q'$.
None of these helps in proving the required statement. In fact, they are asking for a statement when $r$ is rational but not necessarily in $\mathbb{Z}$. Any helpful suggestion is appreciated.
The discriminant of $f$ can be expressed in terms of the resultant, $R$, of $f$ and $f'$, as $$ \Delta(f)=(-1)^{n(n-1)/2}R(f,f') $$ where $n$ is the degree of $f$.
The resultant can be expressed in terms of the product of $f(x)$ evaluated at the zeros of $f'$. Let those zeros be $r$ and $r_2,\dotsc,r_{n-1}$. Then $$ R(f,f')=(-1)^{n(n-1)}n^nf(r)\prod_{i=2}^{n-1}f(r_i) $$ Putting these together, and ignoring the powers of $-1$ which have no effect on divisibility, we have $$ \Delta(f)=n^2f(r)\prod_{i=2}^{n-1}nf(r_i) $$ Now some of the $r_i$, $i=2,\dotsc,n-1$ may be rational, some may be complete sets of conjugates; in any event, they are the roots of $f'/(x-r)$, which is a polynomial of degree $n-2$ with integer coefficients and leading coefficient $n$, so that product is going to be an integer, and we see $f(r)$ divides $\Delta(f)$.