While an explicit function $y(x)$'s arc length $s$ is easily obtained as
$$s = \int \sqrt{1+|y'(x)|^2}\,dx,$$
is there any formula for implicit functions given by $f(x,y) = 0$? One can use the implicit differentiation $y'(x) = -\frac{\partial_y f}{\partial_x f}$ to obtain
$$s = \int\sqrt{1 + |\partial_y f / \partial_x f|^2}\,dx,$$
but that still requires (locally) solving for $y(x)$. Is there any formulation that does not require this, e.g. another implicit equation involving $s$?
Thoughts so far:
One could rewrite $s$ as
$$s = \int |\nabla f|\, |\partial_x f|dx,$$
or symmetrize to
$$s = \int |\nabla f|\, \underbrace{(|\partial_x f|dx + |\partial_y f|dy)}_{(*)}/2$$
where $(*)$ might be strongly related to $|df|$ I guess (though it's not identical due to the $|\cdot|$), but then?
Consider the divergence theorem on the two-dimensional region $\mathcal R = \{(x,y):f(x,y)\le 0\}$ bounded by the curve $\mathcal C = \partial\mathcal R = \{(x,y):f(x,y)=0\}$, $$\iint_{\mathcal R} \nabla\cdot\mathbf v\,\mathrm dA = \oint_{\mathcal C}\mathbf v\cdot\hat{\mathbf n}\,\mathrm d\ell.$$ If we take $\mathbf v=\hat{\mathbf n}=(\nabla f)/\|\nabla f\|$, we have $\mathbf v\cdot\hat{\mathbf n} = 1$, so $$\iint_{\mathcal R} \nabla\cdot\left(\frac{\nabla f}{\|\nabla f\|}\right)\,\mathrm dA = \oint_{\mathcal C}\mathrm d\ell,$$ which is the arc length of the curve.
I don't know if this formula is useful at all, but it does satisfy your requirements.