You will often find in physics people who treat differentials as real numbers and derivative operators as quotients of differentials; this is, however, not quite correct and the only thing that most of the time allows this is the chain rule, mostly in disguise. Recall that the definition of differential $df$ of a function is $$f(a+h)-f(a)=df(a) \cdot h+ \rho(h)$$
and it can be shown that this linear map (df), which acts on vectors to produce scalars, and lives in a space called the dual space $(\mathbb{R}^n)^*$, can be written in a certain basis, called the dual basis, as $\left(\frac{\partial f}{\partial x},\frac{\partial f}{\partial y},\frac{\partial f}{\partial z},...\right)$. This basis is the projection linear map; a map who takes a vector and projects it into some direction. For example, the $(1,0,0)$ vector projected onto the $x$ direction yields $1$. We write this projection as $\pi_i$.The somewhat annoying thing is that, like the cartesian basis of $\mathbb{R}^n$ can be represented as $e_i$, you also have a notation for the dual basis; that is, $dx_i$. Now, many people say that this basis represents a "infinitesimally small change in the direction of $x_i$. But this is also not correct, because $dx_i$ is a linear map, not a number, and when applied to a vector will output a real non-infinitesimal number. The reason for this notation is that because $x_i\longrightarrow x_i$ is a linear function, you have, with a certain abuse of notation, $dx_i\,h=(x_i+h)-x_i+\rho(h)=h$. That is, $\pi_i=dx_i$. This is the reason why you can "treat the differential operator as two variables". Because for a one variable function $\nu(\mu)$, its differential is
$$d\nu=\nu'd\mu$$ or in Leibniz notation
$$d\nu=\frac{d\nu}{d\mu}d\mu$$
So you actually never "multiply or divide differentials". You just recall the definition of differential in the cartesian basis. So the first step is right. The last step is in fact not. When people write $\Delta \mu$ they normally mean a change in $\mu$. But notice what our definition says about the relation between the change in the function and its differential:
$$\Delta \mu=d\mu\Delta \nu + \rho(\Delta \nu)$$ So in fact the differential differs from the small variation by a function $\rho$(which is the function that measures by how much a function fails to be linear; notice that in the case of $\pi_i$, a linear function, $\rho$ is null). However it should also be stated that this function has the only requirement that it be continuous and vanishes at $0$ with $\frac{1}{\Delta \mu}$, that is, $$\lim_{\Delta \mu\rightarrow 0}\frac{\rho}{\Delta \mu}=0$$ So for small variations of the argument, your last step is approximately right. But I see no reason for which you should be writing it in a manifestly wrong way. Expressing the result in terms of the differentials is more suitable, and, might I add, more powerful, because you may proceed by integration to give
$$\frac{d\nu}{\nu}=-\frac{d\mu}{\mu}\iff \log{\frac{|\nu_f|}{|\nu_i|}}=\log{\frac{|\mu_i|}{|\mu_f|}}\iff \mu_f\nu_f=\mu_i\nu_i$$
From whence you conclude that in these conditions, the product of the reduced mass with the rotational frequency is a conserved quantity.
EDIT: Now that I re-read your question, it is quite trivial that this product is conserved because it is equal to $C$. Nonetheless, I still would prefer to use that form of the relation
