Is $f(x,y(x))$ a function of two variables? If so, how is that possible? I don't know how to interpret this function:

Let $D$ be an open subset of $\mathbb{R}^2$ with a continuous function $f:D \rightarrow \mathbb{R}$  and
  \begin{align}
y'(x)=f(x,y(x))
\end{align}
  is a continuous, explicit first-order differential equation defined on $D$.

From $f:\mathbb{R}^2 \rightarrow \mathbb{R}$ I interpret $f$ as a two variable function, $f(x,y)$.
But here $y$ is a function of $x$, so isn't $f(x,y(x))$ a function of one variable, $f:\mathbb{R} \rightarrow \mathbb{R}$?
 A: Note that $f$ is a function of two variables, but the composition $y' = f\circ y$ is a function of one variable, taking as input $x$ and returning $f(x,y(x))$.
A: Why must $f$ be the two-variable function $f(x,y)$? Why can't it be the two-variable function $f(u,v)$ instead?
The point I'm trying to make here is that when we write "$f(x,y)$ is a two-variable function," the $x$ and $y$ in this expression are just placeholders expressing the fact that you need to give $f$ a pair of numbers in order to get a value from it. It is not correct to give these symbols any meaning beyond that in that context; in fact, technically it may be said to be incorrect to say that "$f(x,y)$ is a two-variable function" at all,
because $f(x,y)$ is a notation that denotes the value that $f$ maps the pair $(x,y)$ to, which is a different thing from the function itself.
When we write $y'(x) = f(x,y(x)),$ however, we indicate (somewhat informally) that we have in mind that there is some function named $y$ whose derivative at a value $x$ can be deduced by letting the two-parameter function $f$ map the pair of numbers $x$ and $y(x)$ to the value of the derivative.
A: $f$ is a two variables function: $f:\mathbb{R}^2\rightarrow\mathbb{R}$. So, for a given $(x,y)$ it returns a value $f(x,y)$.
Now, $y'$ is a one variable function: $y':\mathbb{R}\rightarrow\mathbb{R}$. So, for a given $(x)$ it returns a value $y'(x)$.
When one states that $y'(x) = f(x,y(x))$; it means that as long $f$ is evaluated in a point of the curve $(x,y(x))$ it will coincide with $y'(x)$.
What is happening is that we are reducing the domain of $f$, so it will only depend on a single parameter.
A: Writing $f(x, y(x))$ implies that $y$ is a function of $x$, hence $y'(x)$ is only defined on the subset of $\mathbb{R}^2$ that is on the curve $y(x)$. So, even if $f$ is defined on a larger subset writing $y'(x) = f(x, y(x))$ is restricting $f$ to values on the 1-D curve defined by $y(x)$; in this sense it is a 1-D function (whose value is the slope of the curve $y(x)$ at the point $x$).
However, there is nothing stopping $f$ from being defined on all of $\mathbb{R}^2$, those values just won't be relevant to the differential equation.
