Let $D$ domain, $M,N$ $D$-modules with $M$ divisible then $\hom_{D}(M,N)$ is a $D$-module has no torsion
We need to prove $H=\hom_{D}(M,N) \not\subset t(H)$. Let $0 \neq f \in H$. $M$ is divisible, for all $x \in M, n \in \mathbb{Z} \setminus \{0\}$, exists $x' \in M $ such that $nx'=x$. It's enough to prove $\forall d \in D \setminus \{0\}, \exists y \in M, df(y) \neq 0$. I propose $y=x'$, suppose $df(x')=0$ so $ndf(x')=df(x)=0$. But I have problems because $D$ is a domain then not necessarily exists inverse of $d \in D \setminus \{0\}$.
The other idea is try to find a homomorphism no zero such that $\forall d \in D \setminus \{0\}, \forall x \in M, df(x) \neq 0$. But I have no idea how to find it. Could you help me in this problem?
The problem, as you've written it, seems to solve itself if you follow the definitions:
Suppose $f\neq 0$ is any nonzero element of $H$ and $d$ is any nonzero element of $D$.
Then $f(m)\neq 0$ for some $m\in M$.
Since $M$ is divisible, $m=dm'$ for some $m'\in M$. Then $f(dm')=df(m')\neq0$. So $d\cdot f\neq 0$.
We've shown that for every choice of nonzero $f\in H$ and nonzero $d\in D$ that $f\cdot d\neq 0$, so $H$ has no torsion.
Now in the comments above, you seem to disagree with this interpretation, because you say something else needs to be proven. Maybe I have it wrong, but please do reconsider what needs to be proven.