= Solution
For a breathing <displacement> take $\mathbf u_i=q(t)\mathbf r_i$. Every <spring> has equilibrium length $\sqrt2$ in the given length units, so its linear extension is $\sqrt2q$. At the vertex $\mathbf r_i$, the four adjacent equilibrium vertices sum to zero. Therefore the <spring> <force> there is
$$
\mathbf F_i=-kq\sum_{j\sim i}(\mathbf r_i-\mathbf r_j)=-4kq\mathbf r_i.
$$
It remains radial and proportional to the original <displacement> at every vertex, proving that this symmetric <displacement> is a <normal mode>. <Newton's second law> gives
$$
m\ddot q=-4kq,\qquad
\boxed{\omega_b=2\sqrt{\frac km},\qquad f_b=\frac1\pi\sqrt{\frac km}.}
$$
The first is <angular frequency> and the second is cycles per unit time. For $k$ in $\mathrm{N\,m^{-1}}$ and $m$ in $\mathrm{kg}$, these have units $\mathrm{rad\,s^{-1}}$ and $\mathrm{Hz}$ respectively. An <energy> check gives $T=3m\dot q^2$ and $V=12kq^2$, leading to the same ratio $\omega_b^2=24k/(6m)$.
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