= Solution
Take $b$ to be the vector of $f_{ij}$ and absorb $h^{-2}$ into $A$. Its diagonal entry is $-10/(3h^2)$, each axial-neighbour entry is $2/(3h^2)$ and each diagonal-neighbour entry is $1/(6h^2)$. Since these weights agree in both directions, $A$ is a <symmetric matrix>.
Extend a grid vector $u$ by zero to the whole integer lattice. Let $\mathcal E_a$ contain unordered axial-neighbour pairs and $\mathcal E_d$ unordered diagonal-neighbour pairs. Only finitely many differences involving nonzero values contribute. Each lattice vertex has total neighbour weight $4(2/3)+4(1/6)=10/3$, so expanding the squared differences gives
$$
\boxed{u^TAu=-\frac1{h^2}\left[\frac23\sum_{\{v,w\}\in\mathcal E_a}(u_v-u_w)^2+\frac16\sum_{\{v,w\}\in\mathcal E_d}(u_v-u_w)^2\right].}
$$
It is nonpositive. If it vanishes, every axial difference vanishes. The axial lattice is connected and $u$ is zero outside the finite interior square, forcing $u=0$. Thus $u^TAu<0$ for every nonzero $u$: $\boxed{A\text{ is negative definite}}$.
Reordering the grid replaces $A$ by $P^TAP$ for a <permutation matrix> $P$. This preserves symmetry and the sign of the quadratic form, proving the assertion for any ordering. Multiplying the whole system by $h^2$ merely rescales $A$ by a positive number and has the same conclusions.
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