In my book, one way to find the conditional density $f_{Y\mid W}(y \mid w)$ of the random variable $Y = Z/W$ is by taking $W = w$ as a constant and computing $wf_Z(wy)$. I’m wondering intuitively, why this works? I’m having trouble seeing how this is connected to the formal definition $f(y \mid w) = \frac{f(y,w)}{f(w)}$.
Edit:
I posted a sample problem in my book (Mathematical Statistics 7th Edition by Wackerly et. al) below. In this example, $T = \frac{Z}{\sqrt{W/v}}$ is a t-distribution with $v$ df. Hence, $Z$ is a standard normal random variable and $W$ is a $\chi^2$ distributed variable with $v$ df. $Z$ and $W$ are independent.
Here is an intuitive (i.e. non-rigorous) explanation.
If we were dealing with discrete variables and probability masses, then
$$P(Y=y \mid W=w) = P(Z = wy \mid W=w)$$
Nice and obvious. But when dealing with continuous variables and probability densities, a change of variable requires a re-scaling. Intuitively:
$$ \begin{array}{rcl} F_{Y|W}(y|w) = P(Y \le y \mid W=w) &=& P(Z \le wy \mid W=w) = F_{Z|W}(wy|w)\\ {d \over dy} F_{Y|W}(y|w) \times dy &=& {d \over dz} F_{Z|W}(wy|w) \times dz\\ f_{Y|W}(y|w) \times dy &=& f_{Z|W}(wy|w) \times dz\\ f_{Y|W}(y|w) &= &f_{Z|W}(wy|w) \times {dz \over dy}\\ &=& f_{Z|W}(wy|w)\times w\\ &=& w \times f_Z(wy) \,\,\,\,\, \text{if $Z,W$ independent} \end{array} $$
Hopefully your textbook did mention that $Z,W$ are independent! If not, I don't think you can derive $w f_Z(wy)$...
I would not consider the above rigorous, and indeed I don't know enough measure theory to make it rigorous. If you need a more rigorous proof, hopefully someone else can help.