Proving Hadamard's Variational Formula $\dot{\lambda} = -\int_{\partial U(\tau)}\left|\frac{\partial w}{\partial v}\right|^2v\cdot \nu dS$ Setting to this question is the following problem from Evans' book Partial Differential Equations (second edition, page 369):

Let $U(\tau)_{\tau \in \mathbb{R}}$ be a family of smooth bounded domains in $\mathbb{R}^n$, which depend smoothly upon the parameter $\tau \in \mathbb{R}$. Suppose that as $\tau$ changes each point on $\partial U(\tau)$ moves with velocity $v$. For each $\tau$, let us consider eigenvalues $\lambda = \lambda(\tau)$ and the corresponding eigenfunctions $w = w(x, \tau):\begin{cases}-\Delta w = \lambda w&: w \in U(\tau)\\w = 0&: w \in \partial U(\tau)\end{cases}$ normalized such that $||w||_{L^2(U(\tau))} = 1$. Suppose that $\lambda$ and $w$ are smooth functions of $\tau$ and $x$. Show the Hadamard's variational formula:
\begin{equation}
    \begin{aligned}
        \frac{d\lambda}{d\tau} &= -\int_{\partial U(\tau)}\left|\frac{\partial w}{\partial \nu}\right|^2v\cdot \nu dS
    \end{aligned}
\end{equation}

This question has already been asked in Hadamard variational formula Evans chapter 6 problem 15, but I have some questions regarding the given explanation: https://math.stackexchange.com/a/1198921/820472
Namely, 1.) from what does it follow a priori that $\lambda(\tau) = \int_{U}w(-\Delta w)dx = \int_{U}|\nabla w|^2dx$?
Please see the edit.
2.) I am also stuck at trying to conclude the claimed formula $\dot{\lambda} = -\int_{\partial U(\tau)}\left|\frac{\partial w}{\partial v}\right|^2v\cdot \nu dS$. Namely, assuming the claimed equality holds for $\lambda(\tau)$, if I apply the Leibniz formula as suggested, I get $\dot{\lambda}(\tau) = \int_{U(\tau)}(w(-\Delta)w)v \cdot \nu dS(\tau) + \int_{\partial U(\tau)}\frac{\partial}{\partial \tau}(w(-\Delta)w)dx$. Substituting $-\Delta w = \lambda w$ gives
$\dot{\lambda}(\tau) = \int_{U(\tau)}(\lambda w^2)v \cdot \nu dS(\tau) + \int_{\partial U(\tau)}\frac{\partial}{\partial \tau}(\lambda w^2)dx = \int_{U(\tau)}(\lambda w^2)v \cdot \nu dS(\tau) + \int_{\partial U(\tau)}\lambda_\tau w^2 + 2\lambda ww_\tau dx$ But as $w = 0$ on $\partial U(\tau)$, it follows that $\dot{\lambda}(\tau) = \int_{U(\tau)}(\lambda w^2)v \cdot \nu dS(\tau)$
and I have no clue how to proceed in this branch.
The other suggested equality yields $\dot{\lambda}(\tau) = \int_{U(\tau)}(|\nabla w|^2)v \cdot \nu dS(\tau) + \int_{\partial U(\tau)}\frac{\partial}{\partial \tau}(|\nabla w|^2)dx = \int_{U(\tau)}(|\nabla w|^2)v \cdot \nu dS(\tau) + \int_{\partial U(\tau)}\sum_{i=1}^n\frac{\partial}{\partial \tau}\left(\frac{\partial^2}{\partial x_i^2}w\right)^2dx$
where $x := (x_1,\dots,x_n)$
which seems almost the equality we want, provided that we can show $\int_{\partial U(\tau)}\sum_{i=1}^n\frac{\partial}{\partial \tau}\left(\frac{\partial^2}{\partial x_i^2}w\right)^2dx = \int_{\partial U(\tau)}\sum_{i=1}^n2\left(\frac{\partial^2}{\partial x_i^2}w\right)(\frac{\partial^3}{\partial \tau\partial x_i^2}w)dx = 0$. Unfortunately I don't know how to finish the proof.
Edit: I realized just after posting this question that the Hadamard's Variational Formula follows (almost) immediately after applying the Leibniz rule to the equality $\lambda(\tau) = \int_{U}|\nabla w|^2dx$, as $w \equiv 0$ on $\partial U$ implies that $\int_{\partial U(\tau)}\sum_{i=1}^n\frac{\partial}{\partial \tau}\left(\frac{\partial^2}{\partial x_i^2}w\right)^2dx = 0$.
Therefore my renewed questions are the original 1.) and why $\nabla w || \nu \implies \left|\frac{\partial w}{\partial \nu}\right|^2 = |\nabla w|^2$? That is, why if the gradient is parallel to $\nu$ we have that the derivative of $w$ w.r.t. $\nu$ is the squared gradient of $w$?
 A: For 1): Multiplying the equation $-\Delta w = \lambda w$ by $w$ then integrating over $U(\tau)$ gives $$\lambda(\tau) \int_{U(\tau)} w^2 \, dx = - \int_{U(\tau)}w\Delta w \, dx. $$ Then, by assumption, $$ \int_{U(\tau)} w^2 \, dx = \| w\|_{L^2(U(\tau))}^2=1,  $$ so $$\lambda(\tau) = - \int_{U(\tau)}w\Delta w \, dx. $$ Then the equality, $$- \int_{U(\tau)}w\Delta w \, dx = \int_{U(\tau)} \vert \nabla w \vert^2 \, dx $$ is simply integration by parts (using that $w=0$ on $\partial U(\tau)$), see Theorem 3 in Appendix C.2 in Evans.
For 2): We have that $\nabla w$ is parallel to $\nu$ on $\partial U(\tau)$ not perpendicular (i.e. $\nabla w$ is perpendicular to $\partial U(\tau)$) which follows from the fact that $w=0$ on $\partial U(\tau)$. Hence, for each $x\in \partial U(\tau)$, $\nabla w(x)$ is a scalar multiple of $\nu(x)$, that is, $$\nabla w(x) = k \nu (x) $$ for some $k$. To find $k$, we take the dot product with respect to $\nu(x)$ to obtain $$ k = k \nu(x) \cdot \nu(x) =\nabla w(x)\cdot \nu(x) = \frac{\partial w}{\partial \nu}.  $$ Thus, $$\nabla w(x) = \frac{\partial w}{\partial \nu} \nu (x) $$ and this is valid for all $x\in \partial U(\tau)$. Finally, it follows that $$\vert \nabla w(x) \vert^2 = \bigg (\frac{\partial w}{\partial \nu} \bigg )^2 \vert \nu \vert^2 =\bigg (\frac{\partial w}{\partial \nu} \bigg )^2   $$ as required.
