Is this ambiguity in the use of prime notation to denote a derivative? Consider the function $f(x)$ and let $g(x)=f(cx)$.
By the definition of derivative
$$f'(x)=\frac{df(x)}{dx}=\lim\limits_{h \to 0} \frac{f(x+h)-f(x)}{h}\tag{1}$$
and so the definition of $f'(cx)$ is
$$f'(cx)=\frac{df(cx)}{d(cx)}=\lim\limits_{h \to 0} \frac{f(cx+h)-f(cx)}{h}\tag{2}$$
Also
$$g'(x) = \lim\limits_{h \to 0} \frac{g(x+h)-g(x)}{h}$$
$$=\lim\limits_{h \to 0} \frac{f(c(x+h))-f(cx)}{h}$$
$$=\lim\limits_{h \to 0} \frac{f(cx+ch)-f(cx)}{h}$$
$$=c\lim\limits_{h \to 0} \frac{f(cx+ch)-f(cx)}{ch}$$
$$=cf'(cx)$$
Hence
$$g'(x)=cf'(cx)$$
I am not sure if I am seeing an inexistent ambiguity, but $f'(cx)$ seems like ambiguous notation.
As defined above in $(2)$, $f'(cx)$ means the derivative of $f$ relative to $cx$, evaluated at a point we call $cx$, .
Note that this is different than the derivative of $f$ relative to $x$ evaluated at a point $cx$: this derivative is $g'(x)=\frac{df(cx)}{dx}$. In "prime" notation, how do we denote this latter derivative? It would seem to be $f'(cx)$, but I think either this is incorrect, or the definition given in $(2)$ is somehow non-standard.
For example, let $$f(x)=3x^3$$ $$g(x)=f(cx)=3c^3x^3$$
Then
$$f'(x)=\frac{df(x)}{dx}=9x^2$$
$$\left.\frac{df(x)}{dx}\right \vert_{x=cx}=9c^2x^2\ \ (=f'(cx)???)$$
$$g'(x)=\frac{df(cx)}{dx}=c\frac{df(cx)}{d(cx)}=c\cdot f'(cx)=c\cdot 9c^2x^2= 9c^3x^2\tag{3}$$
In this example, $f'(cx)=\frac{df(cx)}{d(cx)}=9c^2x^2$ according to definition I gave in $(2)$.
But perhaps more intuitively, it could also be $f'(cx)= \left.\frac{df(x)}{dx}\right \vert_{x=cx}=9c^2x^2$
Note that both uses of $f'(cx)$ lead to the same result in this example.
Which use of $f'(cx)$ is the "correct" or "standard" one?
 A: I just want to emphasize one thing. You don't take the derivative of a function "relative to something". The derivative of a function $f:\mathbb R \to \mathbb R$ is always taken "with respect to" one thing and one thing only: its entry.
If you want to see how the quantity $f(cx)$ varies when $x$ varies, then you should think of it as defining a new function $g$ by $g(x)=f(cx)$ (because you want to vary $x$) and then differentiating this new function $g$.
My main point is that you have one operation that acts on functions and spits out functions: the derivative. It takes $f$ and gives you $f'$ back. It takes $g$ and gives you $g'$ back. After that, you can evaluate these functions wherever you want. The relation that you correctly found
$$g'(x) = cf'(cx)$$
reads like this: when you evaluate the function $g'$ at the point $x$, you obtain the same number as if you were to evaluate the function $f'$ at the point $cx$ and multiply the result by $c$.
That's it. No "with respect to"s involved.
A: This is just a supplementary part to the nice answer of @KarlSchildkraut. I think it is helpful to consider the complete definition of functions and look at their relationship somewhat more detailed.
The setting more detailed:
We start with a differentiable function
\begin{align*}
&f:\mathbb{R}\to\mathbb{R}\\
&x\mapsto f(x)
\end{align*}
When we now consider
\begin{align*}
g(x)=f(cx)
\end{align*}
we have a function $h$
\begin{align*}
&h:\mathbb{R}\to\mathbb{R}\\
&x\mapsto h(x)=cx\\
\end{align*}
and a function $g$
\begin{align*}
&g:\mathbb{R}\to\mathbb{R}\\
&x\mapsto g(x)=\left(f\circ h\right)(x)\\
\end{align*}
which is the composition of $f$ with $h$. It follows according to the definition of $h$ and $g$ :
\begin{align*}
g(x)=\left(f\circ h\right)(x)=f(h(x))=f(cx)
\end{align*}
Differentiation: Two different situations
As @KarlSchildkraut said there is no ambiguity, but we have different situations instead. On the one hand we have
\begin{align*}
\color{blue}{f^{\prime}(cx)=f^{\prime}(u)\big|_{u=cx}}
\end{align*}
which is the differentiated function $f$ evaluated at a point $u=cx$. On the other hand we have
\begin{align*}
\color{blue}{g^{\prime}(x)}&\color{blue}{=\left(f\circ h\right)^{\prime}(x)}=\left(f(h(x)\right)^{\prime}=\left(f(cx)\right)^{\prime}\\
&=f^{\prime}(h(x))h^{\prime}(x)\\
&\,\,\color{blue}{=f^{\prime}(cx)c}
\end{align*}
which is the differentiation of a composition of functions. So, we have two different situations.
A: This is definitely potentially confusing, but not an ambiguity.
The object written $f$ is a function, which takes an input and gives an output. The object written $f'$ is also a function, which is defined using the function $f$. If $f$ is the function which squares its input (usually written like $f(x)=x^2$), then $f'$ is the function which doubles its input (usually written like $f'(x)=2x$). It's also true that $f'(cx)=2cx$, because $f'(cx)$ does not mean "the derivative of $f(cx)$," it means "the function $f'$ evaluated at an input of $cx$."
Your expression (2) is a correct application of the definition (1), although it would be odd to call it a definition itself. I do not know of a way to write "the derivative of $f(cx)$ as a function of $x$" in prime notation, besides by writing a new function $g(x)$ defined by the rule $g(x)=f(cx)$, from which "the derivative of $f(cx)$ as a function of $x$" is $g'(x)$.
