There is a mathematically serious notion of a functional derivative, but it requires distributions in order to obtain a rigorous foundation. I'll shoot for a hand-wavy view of the situation.
We define the Dirac delta function $\delta$ to be such that $\displaystyle \int_{-\infty}^{+\infty}\delta(x)f(x)dx=f(0)$. Such a thing doesn't technically exist, but if you play by certain rules you can "act" like it does and remain consistent; you'll have to dive more into the theory of distributions if you want a more satisfying explanation.
Additionally, while a function $f$ maps numbers to numbers typically (for example $\mathbb{R}\to\mathbb{R}$), we introduce what's called a functional, which maps functions to numbers. Now we define a derivative
$\frac{\delta \mathcal{F}}{\delta \phi}=\lim_{\epsilon\to0}\frac{\mathcal{F}(\phi(\cdot)+\epsilon\,\delta(\;\cdot\;-y))-\mathcal{F}(\phi(\cdot))}{\epsilon}$
Here $\phi(\cdot)$ is a function, $\mathcal{F}(\;)$ is a functional, $\epsilon$ is restricted to actual numbers, and the derivative here is actually a function of $y$. So if we have a linear functional
$\mathcal{F}(f)=\int_a^b K(x)f(x)dx$
then our derivative is given by
$\lim_{\epsilon\to0}\frac{\int_a^b K(x)[f(x)+\epsilon\,\delta(x-y)]dx-\int_a^b K(x)f(x)dx}{\epsilon}=\lim_{\epsilon\to0}\frac{\int_a^b \epsilon\, K(x)\delta(x-y)dx}{\epsilon}=K(y).$
(Actually it would evaluate to $0$ if $y$ is outside of $[a,b]$.) This is consistent with what the author wrote, with the caveat that there should be a non-$j$ variable on the right-hand side. But as I said, you will need to go at least knee-deep into the theory of distributions in order to fully make sense of these concepts and see how they are rigorously defined. Hope that helps.