Consider $X : p_{X}(x) = \frac{1}{x}\mathbb{1}_{(1,e)}$ and $Y : p_{Y}(x) = 4x^{3}\mathbb{1}(x\in (0,1))$.
We want to find conditional expectation of $\mathbb{E}(\frac{X}{Y}|X-Y)$
My attempt :
$p_{X,Y}(x,y) = \frac{4y^3}{x}\mathbb{1}(1\le x \le e) \mathbb{1}(0\le y\le1)$
So, consider for some $B \in $ B(R) :
$\displaystyle \int_{(X/Y|X-Y) \in B} \frac{4y^{3}}{x}\mathbb{1}(1 \le x\le e) \mathbb{1}(0 \le y \le 1)$, after substitution $x/y = t, x-y = z$ we have $\mathbb{J} = \frac{z}{(1-t)^{2}}$ and so $\displaystyle \int_{(t,z)\in B} \frac{4(t-1)(t-1)^{3}z}{tz(t-1)^2}\mathbb{1}(1 \le \frac{zt}{t-1}\le e) \mathbb{1}(0 \le \frac{z}{t-1} \le 1) = \int_{B} \frac{4(t-1)^2}{tz^3}\mathbb{1}(1 \le \frac{zt}{t-1}\le e) \mathbb{1}(0 \le \frac{z}{t-1} \le 1) dt dz, $
so $p_{(X/Y,X-Y)}(t,z) = \frac{4(t-1)^2}{tz^3}\mathbb{1}(1 \le \frac{zt}{t-1}\le e) \mathbb{1}(0 \le \frac{z}{t-1} \le 1)$
Now we want to find $\displaystyle p_{z} = \int_{\mathbb{R}}\frac{4(t-1)^2}{tz^3}\mathbb{1}(1 \le \frac{zt}{t-1}\le e) \mathbb{1}(0 \le \frac{z}{t-1} \le 1) dt = \int_{\mathbb{R}} \frac{4}{(wz)^{3}(1+w)}\mathbb{1}(1-z\le wz \le e-z)\mathbb{1}(0\le wz\le 1) dw$
There is a first problem : how can I find this integral? I should consider different values of $z$? Am i right ? Any HINTS, cause I have a lot of misunderstandings here.
I can't comment so I am writing this as an answer. Your join density expression is only correct if $X, Y$ are independent. Is this the case?
Otherwise note that $$ \mathbb E \left( \left. {X\over Y} \right| X - Y = a \right) = 1 + \mathbb E \left( \left. {a\over Y} \right| X - Y = a \right) $$
Maybe this can help?