Deriving the Euler-Lagrange Equations for the Chan-Vese Model
Thanks to Luminita Vese, who responded to this question via email. I will post the answer here.  Let $\varphi_\epsilon(x) = \varphi(x) + \epsilon\eta(x)$ for some test function $\eta(x)$.

\begin{eqnarray}
\int_{\Omega} \frac{d}{d\epsilon} \left[ \delta(\varphi_\epsilon(x)) |\nabla (\varphi_\epsilon(x))|\right]dx &amp;=&amp; \int_{\Omega} \left[\frac d{d\epsilon} \delta(\varphi_\epsilon(x))\right] |\nabla (\varphi(x))|+  \delta(\varphi(x)) \left[\frac{d}{d\epsilon}|\nabla (\varphi_\epsilon(x))|\right]dx\\
&amp;=&amp;\int_\Omega \delta'(\varphi(x))\eta(x) |\nabla \varphi(x)| + \delta(\varphi(x))\left[\frac{d}{d\epsilon} \sqrt{\left&lt;\nabla\varphi_\epsilon(x), \nabla\varphi_\epsilon(x)\right&gt;}\right]dx\\
&amp;=&amp;\int_\Omega \delta'(\varphi(x))\eta(x) |\nabla \varphi(x)| + \delta(\varphi(x))\left[ \frac{\frac{d}{d\epsilon}{\left&lt;\nabla\varphi_\epsilon(x), \nabla \varphi_\epsilon(x)\right&gt;}}{2|\nabla(\varphi(x)|}\right]dx\\
&amp;=&amp;\int_\Omega \delta'(\varphi(x))\eta(x) |\nabla \varphi(x)| \\
&amp;\quad&amp;+ \delta(\varphi(x))\left[ \frac{{\left&lt;\nabla\varphi (x), \nabla \eta(x)\right&gt;}}{|\nabla(\varphi(x)|}\right]\\
&amp;=&amp;\int_\Omega \delta'(\varphi(x))\eta(x) |\nabla \varphi(x)| \\
&amp;\quad&amp;+ \frac{{\left&lt;\delta(\varphi(x))\nabla\varphi (x), \nabla \eta(x)\right&gt;}}{|\nabla(\varphi(x)|}dx \quad (*)\\
&amp;=&amp;\int_{\Omega}\delta'(\varphi(x))\eta(x)|\nabla \varphi(x)|\\
&amp;\quad&amp;-\delta(\varphi(x))\mathrm{div}\left(\frac{\nabla \varphi}{|\nabla \varphi|}\right) \eta(x) \\
&amp;\quad&amp;- \delta'(\varphi(x))\frac{|\nabla(\varphi(x))|^2}{|\nabla \varphi(x)|}\eta(x)dx\quad(**)\\
&amp;=&amp;-\int_{\Omega}\delta(\varphi(x))\mathrm{div}\left(\frac{\nabla \varphi}{|\nabla \varphi|}\right) \eta(x) dx
\end{eqnarray}
Above, I am writing $\left&lt; \cdot , \cdot \right&gt;$ to denote the vector inner product, From (*) to (**), Green's first identity was applied as well as the product rule for divergence.