Let $X$, $Y$ be r.v.s with the uniform distribution on the triangle $A$ = $\{(x,y) \in \mathbb{R^2}:0 \leq y \leq x \leq 1\}$. So $d\mu_{(X,Y)}(x,y)= 2\mathbf{1}_A(x,y)dxdy$.
I found that $P(Y/X\leq k)=k$ when $k \in (0,1)$. How do I use this to show that $X$ and $Y/X$ are independent?
It is not strictly necessary to find the distribution of $Y/X$ to show independence, and in fact the distribution will be obvious from the proof. Fix $j$ and $k$, and we want to calculate
$$P\Big( X\le j, \, Y/X \le k\ \Big) = \iint_{A_{j,k}} 2 \, dxdy,$$
where $A_{j,k}:= \big\{ (x,y) \,:\, x\le j, \, y/x \le k \big\}$. Use the change of variables $(x,y) = (u,uv)$ (which we could rewrite as $(u,v) = (x,y/x)$). One has
$$ \begin{bmatrix} \frac{dx}{du} & \frac{dx}{dv} \\ \frac{dy}{du} & \frac{dy}{dv} \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ v & u \end{bmatrix}, $$
so the Jacobian of the transformation is $u$. Note that $(x,y)\in A_{j,k}$ if and only if $0 \le u \le j$, $0 \le v \le k$. Hence,
$$P\Big( X\le j, \, Y/X \le k\ \Big) = 2\int_0^j \int_0^{k/u} u\,dvdu = 2k\int_0^j u\, du = j^2k.$$
Since this is a function of $j$ times a function of $k$, it follows that $X$ and $Y/X$ are independent.