If we have two random variables $X$ and $Y$ ($0<x<y$) and we know the conditional density $f(x\mid y) = \frac{3x^2}{y^3}$, how can we show that $Z = \frac{X}{Y}$ and $Y$ are independent?
Specifically, we don't have $f_Y(y)$ and, as a result, can't say anything about the joint density $f_{X,Y}(x,y)$.
In terms of the definition of independence, I believe that you would want to show that you can express the joint density of $Z$ and $Y$, $f_{Z,Y}(z,y)$, as a separable product $f_Z(z) f_Y(y)$.
The conditional density $f(x|y)$ is related to the joint density $f_{XY}(x,y)$ and the marginal density $f_Y(y)$ by
$$f_{XY}(x,y)= f(x|y)f_{Y}(y)= \frac{3x^2}{y^3}f_{Y}(y).$$
Consider the transformation
$$Z = \frac{X}{Y}, \,\, V = Y,$$
where $0 < Z < 1$ and $V > 0$.
The inverse transformation is
$$X = ZV, \,\, Y = V,$$
with Jacobian
$$J(z,v)=\frac{\partial(X,Y)}{\partial(Z,V)}=\left| \begin{array}{ccc} v & z \\ 0 & 1 \\ \end{array} \right|=v.$$
The joint density of $Z$ and $V$ is
$$f_{ZV}(z,v)= f_{XY}[(X(z,v),Y(z,v)]|J|=f_{XY}(zv,v)v\\=vf(zv|v)f_{Y}(v)= v\frac{3z^2v^2}{v^3}f_{Y}(v)=3z^2f_{Y}(v).$$
Hence, the joint density of $Z=X/Y$ and $Y=V$ is a product of the form
$$f_{Z,Y}(z,y) = 3z^2f_{Y}(y)=f_Z(z)f_Y(y),$$
and $Z = X/Y$ and $Y$ are independent.
