To do it more rigorously (read analysis-y), we have $S^1=\mathbb{R}/2\pi$, a smooth
$$
(\rho,\psi)\colon\mathbb{R}\times S^1\to\mathbb{R}^+\times S^1
$$
such that:
- $\psi(t,-)$ is a diffeomorphism $S^1\to S^1$ for all $t$, say orientation preserving, with inverse $\phi(t,-)$;
- normal deformation flow (*)
\begin{align*}
(\partial_1\rho,\partial_1\psi)
&=v(\psi)\frac{(\rho\partial_2\psi,-\frac1\rho\partial_2\rho)}{\sqrt{(\rho\partial_2\psi)^2+\rho^2(\frac1\rho\partial_2\rho)^2}}\\
&=\left(
\frac{v(\psi)\rho\partial_2\psi}{\sqrt{(\rho\partial_2\psi)^2+(\partial_2\rho)^2}}
,
-\frac{v(\psi)\frac1\rho\partial_2\rho}{\sqrt{(\rho\partial_2\psi)^2+(\partial_2\rho)^2}}\right)
\end{align*}
Then
$$
r(\theta,t):=\rho(t,\phi(t,\theta))
$$
so we calculate
\begin{align*}
\psi(t,\phi(t,\theta))=\theta
&\Rightarrow
\frac{\partial}{\partial t}\psi(t,\phi(t,\theta))=0,\quad
\frac{\partial}{\partial \theta}\psi(t,\phi(t,\theta))=1
\\
&\Rightarrow \partial_1\psi(t,\phi(t,\theta))+\partial_2\psi(t,\phi(t,\theta))\partial_1\phi(t,\theta)=0,\quad
\partial_2\psi(t,\phi(t,\theta))\partial_2\phi(t,\theta)=1\\
&\Rightarrow \partial_1\phi(t,\theta)=-\frac{\partial_1\psi(t,\phi(t,\theta))}{\partial_2\psi(t,\phi(t,\theta))},\quad
\partial_2\phi(t,\theta)=\frac1{\partial_2\psi(t,\phi(t,\theta))}
\end{align*}
hence
$$
\frac{\partial r}{\partial\theta}(\theta,t)=\frac{\partial}{\partial\theta}\rho(t,\phi(t,\theta))=\partial_2\rho(t,\phi(t,\theta))\cdot\partial_2\phi(t,\theta)=\left.\frac{\partial_2\rho}{\partial_2\psi}\right\rvert_{t,\phi(t,\theta)}
$$
and
\begin{align*}
\frac{\partial}{\partial t}r(\theta,t)
&=\partial_1\rho(t,\phi(t,\theta))+\partial_2\rho(t,\phi(t,\theta))\partial_1\phi(t,\theta)\\
&=\partial_1\rho(t,\phi(t,\theta))-\partial_2\rho(t,\phi(t,\theta))\frac{\partial_1\psi(t,\phi(t,\theta))}{\partial_2\psi(t,\phi(t,\theta))}\\
&=\left.\left(\partial_1\rho-\partial_1\psi\frac{\partial_2\rho}{\partial_2\psi}\right)\right\rvert_{(t,\phi(t,\theta))}\\
&=\left.\left(\frac{v(\psi)\rho\partial_2\psi}{\sqrt{(\rho\partial_2\psi)^2+(\partial_2\rho)^2}}+\frac{v(\psi)\frac1\rho\partial_2\rho}{\sqrt{(\rho\partial_2\psi)^2+(\partial_2\rho)^2}}
\frac{\partial_2\rho}{\partial_2\psi}\right)\right\rvert_{(t,\phi(t,\theta))}\\
&=v(\theta)\left.\left(
\frac{(\rho\partial_2\psi)^2+(\partial_2\rho)^2}{\rho\partial_2\psi\sqrt{(\rho\partial_2\psi)^2+(\partial_2\rho)^2}}
\right)\right\rvert_{(t,\phi(t,\theta))}\\
&=v(\theta)\left.\left(
\frac1\rho\sqrt{\rho^2+\left(\frac{\partial_2\rho}{\partial_2\psi}\right)^2}
\right)\right\rvert_{(t,\phi(t,\theta))}\\
&=\frac{v(\theta)}{r(\theta,t)}\,\sqrt{r(\theta,t)^2+\left(\frac{\partial r(\theta,t)}{\partial\theta}\right)^2}.
\end{align*}
(*) The tangent (of the $S^1\to\mathbb{R}^+\times S^1$ at fixed $t$) is $(\partial_2\rho,\partial_2\psi)\in\mathbb{R}^2=T_\rho\mathbb{R}^+\times T_\psi S^1$, taking orthogonal with respect to metric $dr^2+r^2d\theta^2$ on $\mathbb{R}^+\times S^1$ at $(\rho,\psi)$ gives $\pm(\rho\partial_2\psi,-\frac1\rho\partial_2\rho)$. Since we assume $\psi(t,-)$ is orientation-preserving we choose the + sign to have the first component positive.
Best Answer
Evans's hint is the Leibniz rule: $$ \frac{d}{d\tau}\int_{U(\tau)} F(x,\tau) dx = \int_{\partial U(\tau)} F(x,\tau) \, v\cdot \nu \, dS(\tau) + \int_{U} F_\tau(x,\tau) \, dx,$$ where $v$ is the velocity of the boundary $\partial U$ with respect to $\tau$ and $\nu$ is the outward unit normal. (See here or here for details, but the details aren't too important for the problem.)
Based on the hint and the variational formula, we should probably write $\lambda(\tau)$ as an integral and differentiate with respect to $\tau$. This can be accomplished in two ways: $$ \lambda(\tau) = \int_U w(-\Delta w) \, dx = \int_U |\nabla w|^2 \, dx. $$
Now apply the Leibniz rule to the above equation. Differentiating the first equality gives you some useful information when dealing with the second equality. Use the identity $-\Delta w = \lambda w$ and multivariable integration by parts (see Appendix C in Evans, for example) as needed. The answer should come out.
Another note is that since $w$ vanishes on $\partial U$, the gradient $\nabla w$ is perpendicular to $\partial U$ and thus parallel to the unit normal $\nu$. Therefore, the $|\frac{\partial w}{\partial \nu}|^2$ in Hadamard's formula is just $|\nabla w|^2$.
More about the formula: Hadamard's Formula Inside and Out