Can anyone help me with the determinant of this matrix?
$ \mathbf{J}=\begin{bmatrix}\frac{x_{1}}{y_{1}} & \frac{x_{1}}{y_{2}} & \frac{x_{1}}{y_{3}} & \cdots & \frac{x_{1}}{y_{d-1}} & \frac{x_{1}}{y_{d}}\\ \frac{x_{2}}{y_{1}} & \frac{x_{2}}{y_{2}} & \frac{x_{2}}{y_{3}} & \cdots & \frac{x_{2}}{y_{d-1}} & \frac{-x_{2}}{1-y_{d}}\\ \frac{x_{3}}{y_{1}} & \frac{x_{3}}{y_{2}} & \frac{x_{3}}{y_{3}} & & \frac{-x_{3}}{1-y_{d-1}} & 0\\ \vdots & \vdots & & \cdot^{\cdot^{\cdot}} & 0 & \vdots\\ \frac{x_{d-1}}{y_{1}} & \frac{x_{d-1}}{y_{2}} & \frac{-x_{d-1}}{1-y_{3}} & \cdot^{\cdot^{\cdot}} & 0 & 0\\ \frac{x_{d}}{y_{1}} & \frac{-x_{d}}{1-y_{2}} & 0 & \cdots & 0 & 0 \end{bmatrix}. $
It must be simple due to the triangular structure but I don't get it....
Hints :
A first step : factorization of $J$ (see the remark by @saulspatz) :
$$J=diag(x_1,x_2,\cdots,x_{d-1},x_d) \times K \times diag(y_1,y_2,\cdots,y_{d-1},y_d)$$
where $K$ is a matrix with non-zero entries at the same place as in $J$ but with every $x_k/y_k$ replaced by $1$ and every $-x_k/(1-y_k)$ replaced by $-y_k/(1-y_k)$.
Let us now define matrix $R$ as the "reversal" matrix with entries $0$ except on the anti-diagonal where $R_{k,n-k}=1$.
Please note that $\det(R)=(-1)^{n-1}$.
$H:=KR$ has the same columns as $K$ but in a reversed order, making it a so-called "upper Hessenberg matrix" (that can be described as a upper triangular to which a lower diagonal has been added) (https://en.wikipedia.org/wiki/Hessenberg_matrix).
It remains now to compute $\det(H)$, using one of the algorithms for the computation of the determinant of a Hessenberg matrix : see (journal.pmf.ni.ac.rs/filomat/index.php/filomat/article/download/4595/2309) or (https://math.stackexchange.com/q/3206872).