It is easy to show that a hermitian operator must have real eigenvalues.
It is also easy to show that the operator $-i\dfrac{\partial}{\partial x}$ is hermitian.
But from a computational perspective, I feel like it is possible to just propose working counter-examples. Specifically, the following eigenvector problem, $$ -i\dfrac{\partial}{\partial x}f(x) = \lambda f(x) $$
appears to be solved by $f(x) = e^{-3x}$ with $\lambda = 3i \not\in \mathbb{R}$. We can verify by substitution: \begin{align} -i\dfrac{\partial}{\partial x}e^{-3x} &= 3ie^{-3x}\\ \\ 3ie^{-3x} &= 3ie^{-3x} \end{align}
This imaginary eigenvalue contradicts the proofs I linked to. I must be doing something illogical here, or misunderstanding the proofs.
What an operator is depends on which space it is defined on. Your link shows that $-i\dfrac{\partial}{\partial x}$ is Hermitian on $L^2$, $e^{-3x}$ is not in $L^2$, so it is not a counterexample. Doing this naively one can find an "eigenvector" of $-i\dfrac{\partial}{\partial x}$ for any complex number as an eigenvalue, namely $e^{ax}$ with a suitable $a$.