In the linear sigma model, the Lagrangian is given by

$ \mathcal{L} = \frac{1}{2}\sum_{i=1}^{N} \left(\partial_\mu\phi^i\right)\left(\partial^\mu\phi^i\right) +\frac{1}{2}\mu^2\sum_{i=1}^{N}\left(\phi^i\right)^2-\frac{\lambda}{4}\left(\sum_{i=1}^{N}\left(\phi^i\right)^2\right)^2 $

(for example see Peskin & Schroeder page 349).

When perturbatively computing the effective action for this Lagrangian the derivative $ \frac{\delta^2\mathcal{L}}{\delta\phi^k(x)\delta\phi^l(x)} $ needs to be computed. (for instance, Eq. (11.67) in P&S):

$$

\frac{\delta^2\mathcal{L}}{\delta\phi^k(x)\delta\phi^l(x)} ~=~ -\partial^2\delta^{kl} +\mu^2\delta^{kl}-\lambda\left[\phi^i\phi^i\delta^{kl}+2\phi^k\phi^l\right].\tag{11.67}$$

My question: where are two delta functions?

If you don't understand why there need them, I write full calculation:

\begin{eqnarray}

\frac{\delta^{2} \mathcal{L} \left[\phi\right]}{\delta\phi^{a}\left(x\right)\delta\phi^{b}\left(y\right)}&=&\frac{\delta^{2}}{\delta\phi^{a}\left(x\right)\delta\phi^{b}\left(y\right)}\left\{ \frac{1}{2}\sum_{i=1}^{N}\left(\partial_{\mu z}\phi^{i}\left(z\right)\right)\left(\partial^{\mu}\,_{z}\phi^{i}\left(z\right)\right)+...\right\} \\&\stackrel{}{=}&\frac{\delta}{\delta\phi^{a}\left(x\right)} \left\{ \sum_{i=1}^{N}\left(\left(\partial^{\mu}\,_{z}\phi^{i}\left(z\right)\right)\left(\partial_{\mu}\,_{z}\frac{\delta}{\delta\phi^{b}\left(y\right)}\phi^{i}\left(z\right)\right)\right)+...\right\} \\&=&\frac{\delta}{\delta\phi^{a}\left(x\right)} z\left\{ \sum_{i=1}^{N}\left(\left(\partial^{\mu}\,_{z}\phi^{i}\left(z\right)\right)\left(\partial_{\mu}\,_{z}\delta^{ib}\delta\left(z-y\right)\right)\right)+...\right\} \\&=&\frac{\delta}{\delta\phi^{a}\left(x\right)} \left\{ \left(\partial^{\mu}\,_{z}\phi^{b}\left(z\right)\right)\left(\partial_{\mu}\,_{z}\delta\left(z-y\right)\right)+...\right\} \\&=& \left(\partial^{\mu}\,_{z}\delta^{ab}\delta\left(x-z\right)\right)\left(\partial_{\mu}\,_{z}\delta\left(z-y\right)\right)+... \\&=& -\delta^{ab}\left(\partial_{\mu}\,_{z}\partial^{\mu}\,_{z}\delta\left(x-z\right)\right)\delta\left(z-y\right)+... \\&

\end{eqnarray}

You may see two delta functions there.