Notation of derivatives in field theory Some textbooks write
$$
\frac{\delta F_{\mu\nu}}{\delta(\partial_\sigma A_\kappa)}
$$
which sort of implies the derivative of a functional. Some other textbooks write
$$
\frac{\partial F_{\mu\nu}}{\partial(\partial_\sigma A_\kappa)}
$$
which one would usually read as a partial derivative. Since the Fieldtensor $F_{\mu\nu}$ to my understanding is not a functional, why these notation?
And if $F_{\mu\nu}$ is a functional, why is that ?
 A: There are two notions which tend to be a bit confused in Lagrangian mechanics for fields. 
First, there is the functional derivative. As the name implies, this is a derivative for functionals, such as the action functional. Functionals are maps from function spaces to $\mathbb{R}$ (or some other field). If we consider, say, a scalar field $\phi$, the action functional is
\begin{eqnarray}
S : C_0^\infty(\mathbb{R}) &\to& \mathbb{R}\\
\phi &\mapsto& S[\phi] = \int \mathcal{L}(\phi, \partial \phi) d^4x
\end{eqnarray}
where we consider $\phi$ to belong to $C_0^\infty(\mathbb{R^3})$ here, the functions of compact support in space.
The functional derivative of $S$ can be defined a few ways, but it's usually something like a Gâteau derivative. As we're dealing with a function space, which is usually infinite-dimensional (and more to the point, not one-dimensional), we need to define a direction for it. That direction is a vector in the function space, ie a function itself. If we pick a function $h \in C_0^\infty(\mathbb{R^3})$, the functional derivative is basically defined the same way as any directional derivative.
\begin{eqnarray}
\frac{\delta_h S}{\delta \phi} = \lim_{\varepsilon \to 0} \frac{S[\phi + \varepsilon h] - S[\phi]}{\varepsilon}
\end{eqnarray}
Fortunately, in the case of Lagrangian mechanics, the direction doesn't matter. Given our action functional, for any function $h$, this derivative will be zero at any function $\phi$ which satisfies the Euler-Lagrange equation (ignoring any issues related to gauge or boundaries, to keep things simple) : 
\begin{eqnarray}
\frac{\partial L}{\partial \phi} - \nabla \frac{\partial L}{\partial(\partial \phi)} = 0
\end{eqnarray}
The derivatives here are, roughly speaking, your usual derivatives. To do this in a bit more detail, the Lagrangian here is a function of the form (to simplify)
\begin{eqnarray}
L : \mathbb{R} \times \mathbb{R}^4 &\to& \mathbb{R}\\
(f, v) &\mapsto& L(f, v)
\end{eqnarray}
I wrote the value of the field as $f$ and its derivatives as $v$ to drive home the notion that this is just a regular function at a point. What we do is to compose it with $\phi$ to obtain a function from the function space to $\mathbb{R}$, ie at a point $x$, $f = \phi(x)$, $v = \partial \phi(x)$. Then the derivatives are simply
\begin{eqnarray}
\frac{\partial L}{\partial \phi} &=& \frac{\partial L(f, v)}{\partial f}\Bigr|_{f = \phi(x), v = \partial \phi(x)}
\end{eqnarray}
Physicists do tend to denote this with a delta, but it is very much just an ordinary derivative, in this sense.
