Let $D\subseteq\mathbb{R}^2$ be some domain. Let us assume for the moment that it is compact. If it helps you then you may assume it is $[-L,L]^2$ for some $L>0$, or a closed disc if you prefer.
Let $f:D\to\mathbb{R}$ be a fixed given function that is as regular as you want and might obey some more conditions (like never taking the value zero).
Define an operator $A$ on $L^2(D\to\mathbb{R}^2)$ (I am not actually sure this is the appropriate Hilbert space--in particular, I am not sure if the Hilbert space should be real or complex) as $$A := \frac{1}{f}\begin{bmatrix}-\partial_2 & \partial_1 \\ \partial_1 & \partial_2\end{bmatrix}\,.$$
Here, it is meant that $\frac{1}{f}$ would work as a multiplication operator after having performed the derivatives.
The boundary conditions for this operator are zero i.e., on $\partial D$ the eigenfunctions should vanish.
How can one find the eigenvalues $\lambda_i$ and eigenvectors $\begin{bmatrix}\varphi_{1i} \\\varphi_{2i}\end{bmatrix}\in L^2(D\to\mathbb{R}^2)$ of $A$ (I assume the spectrum is discrete because the domain is compact)? That is, solve the equation $$ A \begin{bmatrix}\varphi_{1i} \\\varphi_{2i}\end{bmatrix} = \lambda_i \begin{bmatrix}\varphi_{1i} \\\varphi_{2i}\end{bmatrix}\,. $$
It is clear that $A$ is a multiple of a Dirac operator. If we multiply it by $-\mathrm{i}$ then we get a self-adjoint operator on $L^2(D)\otimes\mathbb{C}^2$ (so possibly changing the Hilbert space?) given by $$-\mathrm{i}A = \frac{1}{f}(-P_2\otimes\sigma_3 + P_1\otimes\sigma_1)$$ where $P_j \equiv -\mathrm{i}\partial_j$ is self-adjoint; $-\mathrm{i}A$ is manifestly self-adjoint and hence has real eigenvalues. It must also have spectrum symmetric about zero as a Dirac operator.
However, how to actually find its spectrum? For starters, what if we assume $f$ is harmonic (the real part of an analytic function?)? If $f=1$ then we can work if Fourier space and need to diagonalize $$ \begin{bmatrix}-p_2 & p_1 \\ p_1 & p_2\end{bmatrix} $$ which has eigenvalues $\pm \sqrt{p_1^2+p_2^2}$ with $p$ discrete so as to obey the boundary conditions.