What exactly is the reasoning for why $\frac{d}{dx}u(x, Ce^x) = \frac{\partial{u}}{\partial{x}} + Ce^x \frac{\partial{u}}{\partial{y}}$? I am currently studying the textbook Partial Differential Equations – An introduction, second edition, by Walter A. Strauss. The section The Variable Coefficient Equation of chapter 1 says the following:

The equation
$$u_x + y u_y = 0 \label{4}\tag{4}$$
is linear and homogeneous but has a variable coefficient ($y$). We shall illustrate for equation \eqref{4} how to use the geometric method somewhat like Example 1. The PDE \eqref{4}  itself asserts that the directional derivative in the direction of the vector $(1, y)$ is zero. The curves in the $xy$ plane with $(1, y)$ as tangent vectors have slopes $y$ (see Figure 3). Their equations are
$$\dfrac{dy}{dx} = \dfrac{y}{1} \label{5}\tag{5}$$
This ODE has the solutions
$$y = Ce^x \label{6}\tag{6}$$
These curves are called the characteristic curves of the PDE \eqref{4} . As $C$ is changed, the curves fill out the $xy$ plane perfectly without intersecting. On each of the curves $u(x, y)$ is a constant because
$$\dfrac{d}{dx}u(x, Ce^x) = \dfrac{\partial{u}}{\partial{x}} + Ce^x \dfrac{\partial{u}}{\partial{y}} = u_x + yu_y = 0.$$


What exactly is the reasoning for why $\dfrac{d}{dx}u(x, Ce^x) = \dfrac{\partial{u}}{\partial{x}} + Ce^x \dfrac{\partial{u}}{\partial{y}}$? This seems to be an application of the chain rule, but I don't understand the reasoning behind why the chain rule is appropriate for this case, or for how it is applied.
 A: Let $g:\mathbb R\to\mathbb R^2$ be given by $g(x)=(x,Ce^x)$. Then
$$g'(x) = (1,Ce^x).$$
Now, the notation
$$\frac{d}{dx}u(x,Ce^x)=\dots$$
is a shorthand for
$$(u\circ g)'(x)=\dots.$$
In this case, the chain rule is appropiate because you are differentiating the composition of two functions: $u$ and $g$.
Now the chain rule says
$$(u\circ g)' = (\nabla u \circ g )\cdot g'$$
Evaluating at $x$
$$\begin{align*}
(u\circ g)'(x)
&= \nabla u(g(x))\cdot g'(x) \\
&= (\partial_1 u(x,Ce^x),\partial_2u(x,Ce^x))\cdot(1,Ce^x) \\
&= \partial_1 u(x,Ce^x) + \partial_2u(x,Ce^x)Ce^x
\end{align*}$$
Where $\partial_1u$ and $\partial_2u$ are the partial derivatives of $u$ with respect to its first and second entries.
Please note that, usually, people tend to write things like
$$\frac{\partial u}{\partial x}, \frac{\partial u}{\partial y}$$
or even
$$u_x, u_y$$
to mean $\partial_1u$, and $\partial_2u$, respectively.
In this case, in the member
$$\dots=\frac{\partial u}{\partial x} + Ce^x\frac{\partial u}{\partial y}=\dots$$
the symbols $\frac{\partial u}{\partial x}$ and $\frac{\partial u}{\partial y}$ are used to mean $\partial_1u$ and $\partial_2u$ evaluated at $(x,Ce^x)$. All this without saying. Furthermore, in the member
$$\dots=u_x+yu_y$$
the symbols $u_x$ and $u_y$ are used to mean $\partial_1u$ and $\partial_2u$ evaluated at $(x,Ce^x)$, while the symbol $y$ is used to mean the second entry of $g(x)=(x,Ce^x)$, which is $Ce^x$.
Hope it helps.
