This is typically an exercise for a polynomial ring $R$.
Let $\mathbf{k}$ be an integral domain and let $\mathbf{k}(x)$ denote the field of rational functions of $x$ with coefficients in $\mathbf{k}$ considered as an $\mathbf{k}[x]$-module. Then the dual module $\mathbf{k}(x)^\ast = \mathrm{Hom}_{\mathbf{k}[x]}\big(\mathbf{k}(x),\mathbf{k}[x]\big)$ is the zero module. We can see this by taking $\phi \in \mathbf{k}(x)^\ast$, supposing $\phi(1)$ has degree $n$, and noticing that $$ \phi(1) = \phi\left(\frac{x^{n+1}}{x^{n+1}}\right) = x^{n+1}\phi\left(\frac{1}{x^{n+1}}\right)\,, $$ so $\phi(1)$ has a degree $n+1$ factor. This is a contradiction unless $\phi(1) = 0$.
This argument relies on $R=\mathbf{k}[x]$ being an $\mathbf{N}$-graded ring though. But this must be true more generally, right? For which integral domains $R$ with field of fractions $\mathrm{Frac}(R)$ do we still have $$ \mathrm{Frac}(R)^\ast = \mathrm{Hom}_R\big(\mathrm{Frac}(R),R\big) = 0\,? $$
If $R$ is an integral domain other than a field, then the dual module of the field of fractions of $R$ always vanishes. Indeed, if $f$ is any $R$-linear map from $\mathrm{Frac}(R)$ to $R$, then we must have that $\forall r \in R \forall s \in R \backslash \{0\} \space sf(\frac{r}{s})=f(r)$, so the image of $f$ must be contained in the intersection of all the nonzero principal ideals of $R$. But it can easily be shown that if $r$ is any nonunit in $R$, then any nonzero element $s$ in $R$ cannot lie in the principal ideal generated by $rs$, so $f$ must vanish everywhere.