What does the following notation mean? (ODEs) I saw this notation many times, but I don't understand why the $y$ variable is missing in the first term of the first equation below.
$$
\frac{\mathrm{d}y(x)}{\mathrm{d}x} = f(x,y)
$$
It just mean:
$$
\frac{\mathrm{d}y(x,y)}{\mathrm{d}x} = f(x,y)
$$
?

Tell me if I'm wrong
In my understanding this is a linear ODE of first order in which $y$ is the unknown function and both of them (the derivate and the unknown one) have just a single variable.
$$
y'(x) = f(x)
$$
A simple example
$$
\begin{align}
y'(x) &= 2x\\
\int y'(x)\!\mathop{}\mathrm{d}x &= \int 2x\!\mathop{}\mathrm{d}x\\
y(x)&= x^2+\mathrm{C}\\
\end{align}
$$
Is this case:
$$
\dfrac{\mathrm dy}{\mathrm dx}(x)=f(x,y(x))
$$
would it be something like this?
$$
\begin{align}
y'(x,y) &= 2x + 2y\\
\iint y'(x,y)\!\mathop{}\mathrm{d}x\!\mathop{}\mathrm{d}y &= \iint 2x + 2y\!\mathop{}\mathrm{d}x\!\mathop{}\mathrm{d}y\\
y(x,y) &= x^2 + y^2+\mathrm{C}\\
\end{align}
$$

For Christian Blatter

This is what I would expected to be your exaplanation. Can you tell me
  why my version is wrong?

When dealing with ODEs for the first time we are given a function $f:\ (x,y)\mapsto f(x,y)$ defined in some region $\Omega$ of the $(x,y)$-plane. For each $(x,y)\in\Omega$ the value $f(x,y)$ is to be interpreted as a slope assigned to the point $(x,y)$. Therefore we are given a field of "line elements" of various slopes covering $\Omega$.
Given this "slope field" we are interested in curves $$\gamma:\ x\mapsto \gamma(x)=\bigl(x,\gamma(x)\bigr)\qquad(a<x<b)$$ lying in $\Omega$ that have for each of their points $\bigl(x,\gamma(x)\bigr)$ the slope $f\bigl(x,\gamma(x)\bigr)$ prescribed there. This means that we should have
$$\gamma'(x)\equiv f\bigl(x,\gamma(x)\bigr)\qquad(a<x<b)\ .\tag{1}$$
 A: Your difficulty stems from the use of the letter $y$ for two different purposes: (a) as coordinate variable in the $(x,y)$-plane, and (b) as variable for (unknown) functions $x\mapsto y(x)$ whose graphs are lying in the $(x,y)$-plane.
When dealing with ODEs for the first time we are given a function $f:\ (x,y)\mapsto f(x,y)$ defined in some region $\Omega$ of the $(x,y)$-plane. For each $(x,y)\in\Omega$ the value $f(x,y)$ is to be interpreted as a slope assigned to the point $(x,y)$. Therefore we are given a field of "line elements" of various slopes covering $\Omega$.
Given this "slope field" we are interested in curves $$\gamma:\ y=\phi(x)\qquad(a<x<b)$$ lying in $\Omega$ that have at each of their points  $\bigl(x,\phi(x)\bigr)$ the slope $f\bigl(x,\phi(x))$ prescribed there. Note that these  curves $\gamma$ are considered as graphs of unknown functions $\phi:\>x\mapsto y=\phi(x)$. The condition about the slopes means that we should have
$$\phi'(x)\equiv f\bigl(x,\phi(x)\bigr)\qquad(a<x<b)\ .\tag{1}$$
Now since centuries it is common to denote the unknown functions in such a problem not by $\phi$, or some other letter, but by the same letter as "the coordinate in which they take values", i.e., $y$ in our case. In view of $(1)$ this means that we are looking for functions $x\mapsto y(x)$ such that
$$y'(x)\equiv f\bigl(x,y(x)\bigr)\qquad(a<x<b)\ .\tag{2}$$
The condition $(2)$ is then further abbreviated to
$$y'=f(x,y)\qquad\bigl((x,y)\in\Omega\bigr)\ .\tag{3}$$
The ODE $(3)$ encodes the ideas described here and condenses them into four letters plus some extra tokens. 
Update concerning your "understanding": The unknown functions are functions of one variable $x$, whether one writes them as $\phi:\>x\mapsto y=\phi(x)$ or as $x\mapsto y(x)$. These functions have derivatives with respect to $x$ that are again functions of the one variable $x$. Therefore expressions of the form $y(x,y)$ or ${dy(x,y)\over dx}$ don't make sense. On the other hand the so-called right side of the ODE, a given function $f(x,y)$ defining the desired slope at each point $(x,y)\in\Omega$ is (in general) a function of the two variables $x$ and $y$. In some special cases, as with the super-simple ODE $y'=f(x)$, or with the ODE $y'=g(y)$ only one variable appears in $f$. The interpretation of this phenomenon is the following: In the first case the assigned slope is constant along vertical lines, and in the second case it is constant along horizontal lines.
A: The notation $\dfrac{\mathrm dy(x)}{\mathrm dx}$ is short for $\dfrac{\mathrm dy}{\mathrm dx}(x)$ or $y'(x)$, if you prefer.
In this context, the equality $\dfrac{\mathrm{d}y(x)}{\mathrm{d}x} = f(x,y)$ should be read as $\dfrac{\mathrm dy}{\mathrm dx}(x)=f(x,y(x))$.
As for your last example, you got the wrong idea, $y$ is a function whose domain is a subset of $\mathbb R$, so $y(x,y)$ doesn't make sense.
Correct would be to consider the function $f\colon \mathbb R^2\to \mathbb R, (x,y)\mapsto 2x+2y$ and to consider the ODE $y'(x)=f(x,y(x))$. Note that in $f\colon \mathbb R^2\to \mathbb R, (x,\color{red} y)\mapsto 2x+2\color{red} y$ the variable $\color{red} y$ is bounded, it's not a function, it's a variable. If you wish you can rewrite the function as $f\colon \mathbb R^2\to \mathbb R, (x,u)\mapsto 2x+2u$ and then you can consider the ODE $y'(x)=f(x,y(x))$ without any ambiguity.
