Comment to the question (v2): P&S is using the notation of a 'same-spacetime' functional derivative. To illustrate this notation, let us for simplicity stay within first variations, and leave it to the reader to generalize to higher-order variations.
I) First of all, functional/variational derivatives should not be confused with partial derivatives. In practice, from an operational point of view (if we are not worried about mathematical details about existence and boundary terms), all we need to know is the following rules:
The formula
$$\tag{A} \frac{\delta \phi^{\beta}(y)}{\delta\phi^{\alpha}(x)}
~=~\delta^{\beta}_{\alpha}~\delta^n(x-y), $$
where $n$ is the spacetime dimension.
Appropriate generalizations of elementary rules in calculus, such as, e.g., the chain rule, integration by parts, commutativity of derivatives, and the Dirac delta distribution.
For instance, by these rules 1 & 2, we have that
$$ \frac{\delta}{\delta\phi^{\beta}(y)} \frac{\partial}{\partial x^{\mu_1}}\ldots \frac{\partial}{\partial x^{\mu_r}}\phi^{\alpha}(x)
~=~ \frac{\partial}{\partial x^{\mu_1}}\ldots \frac{\partial}{\partial x^{\mu_r}} \frac{\delta}{\delta\phi^{\beta}(y)}\phi^{\alpha}(x)$$
$$\tag{B}~\stackrel{(A)}{=}~\delta_{\beta}^{\alpha}~\frac{\partial}{\partial x^{\mu_1}}\ldots \frac{\partial}{\partial x^{\mu_r}}\delta^n(x-y). $$
Similarly, by rules 1 & 2, we can deduce that the action
$$\tag{C}S~=~\int d^nx ~{\cal L}(x) , \qquad {\cal L}(x)\equiv {\cal L}(\phi(x), \partial \phi(x), \ldots, x),$$
has the Euler-Lagrange expression as its functional derivative
$$ \tag{D}\frac{\delta S}{\delta\phi^{\alpha} (x)}~=~ \frac{\partial{\cal L}(x) }{\partial\phi^{\alpha} (x)} - d_{\mu} \left(\frac{\partial{\cal L}(x) }{\partial\partial_{\mu}\phi^{\alpha} (x)} \right)+\ldots.$$
The ellipsis $\ldots$ in eqs. (C) and (D) denotes possible contributions from higher-order spacetime derivatives.
II) From formula (A) it becomes clear that it does not makes sense to
consider the functional derivative $\frac{\delta {\cal L}(x)}{\delta\phi^{\alpha} (x)}$ wrt. the same spacetime argument $x$, because that would lead to infinities, cf. $\delta^n(0)=\infty$. Nevertheless, it is tempting to introduce the notation of a 'same-spacetime' functional derivative
$$\tag{E}\frac{\delta {\cal L}(x)}{\delta\phi^{\alpha} (x)}~:=~ \frac{\partial{\cal L}(x) }{\partial\phi^{\alpha} (x)} - d_{\mu} \left(\frac{\partial{\cal L}(x) }{\partial\partial_{\mu}\phi^{\alpha} (x)} \right)+\ldots.$$
We stress that eq. (E) is only a notational definition. It becomes meaningless if we try to interpret the lhs. of eq. (E) using the above rules 1 & 2.
III) Similarly, P&S talk about second-order 'same-spacetime' functional derivative
$$\tag{F}\frac{\delta^2 {\cal L}(x)}{\delta\phi^{\alpha} (x)\delta\phi^{\beta}(x)}.$$
We recommend to first work out the ordinary second-order functional derivative
$$\tag{G}\frac{\delta^2 S}{\delta\phi^{\alpha} (x)\delta\phi^{\beta}(y)}$$
using rules 1 & 2. Then it should be fairly straightforward to translate (G) into the 'same-spacetime' functional derivative language (F), if needed. [In particluar, eq. (G) contains a $\delta^n(x-y)$ while eq. (F) does not.]
IV) Finally we should mention that in field theory one often suppresses the spacetime indices $x,y,\ldots$, by using DeWitt's condensed notation.
Best Answer
Simply read the diagonal N×N matrix $$ -\partial^2\delta^{ij}+\mu^2\delta^{ij} - \lambda[ v^2 \delta^{ij} + 2\phi_{\mathrm{cl}}^i\phi_{\mathrm{cl}}^j], \tag{11.67}$$ where, to allay your confusion, I call $\phi_{cl}=v$, so $(\phi_{\mathrm{cl}}^k)^2=v^2$.
The non-gradient part of this diagonal matrix is, for $i,j= 1...,N-1$, $$(\mu^2-\lambda v^2, \mu^2-\lambda v^2,...,\mu^2-\lambda v^2),$$ whereas the last entry (only!) in the diagonal is $$ \mu^2-\lambda v^2-2\lambda v^2= \mu^2-3\lambda v^2. $$