Expand the determinant $D=\det k(x_j,y_k)$ along the first row. This shows that as a function of $x=x_1$, it is of the form $$ D(x) = \sum_{j=1}^n \frac{c_j}{1-y_j x} , $$ with $c_j$ independent of $x=x_1$ (and of $y_j$, but that doesn't matter here).

This rational function has $n$ poles at $1/y_1,\ldots, 1/y_n$. On the other hand, clearly $D(x_2)=\ldots =D(x_n)=D(\infty)=0$, and since this is a total of $n$ zeros, we have found all of them. Thus $D\not= 0$ under your assumptions.

Small detail added later: If $y_j=0$ for some $j$, then we need to slightly modify the argument: now we have $n-1$ poles, and also $n-1$ zeros since now $D(\infty)=c_j\not= 0$. (This last fact we could also have obtained directly from an inductive argument since the $c_j$ are determinants of $(n-1)\times (n-1)$ submatrices of the same type.)

Expand the determinant $D=\det k(x_j,y_k)$ along the first row. This shows that as a function of $x=x_1$, it is of the form $$ D(x) = \sum_{j=1}^n \frac{c_j}{1-y_j x} , $$ with $c_j$ independent of $x=x_1$ (and of $y_j$, but that doesn't matter here).

This rational function has $n$ poles at $1/y_1,\ldots, 1/y_n$. On the other hand, clearly $D(x_2)=\ldots =D(x_n)=D(\infty)=0$, and since this is a total of $n$ zeros, we have found all of them. Thus $D\not= 0$ under your assumptions.

Small details added later: If $y_j=0$ for some $j$, then we need to slightly modify the argument: now we have $n-1$ poles, and also $n-1$ zeros since now $D(\infty)=c_j\not= 0$. This last fact we could also have obtained directly from an inductive argument since the $c_j$ are determinants of $(n-1)\times (n-1)$ submatrices of the same type. Strictly speaking, we actually need something along these lines no matter what since the argument obviously breaks down if we could have $c_1=\ldots =c_n=0$.

Expand the determinant $D=\det k(x_j,y_k)$ along the first row. This shows that as a function of $x=x_1$, it is of the form $$ D(x) = \sum_{j=1}^n \frac{c_j}{1-y_j x} , $$ with $c_j$ independent of $x=x_1$ (and of $y_j$, but that doesn't matter here).

This rational function has $n$ poles at $1/y_1,\ldots, 1/y_n$. On the other hand, clearly $D(x_2)=\ldots =D(x_n)=D(\infty)=0$, and since this is a total of $n$ zeros, we have found all of them. Thus $D\not= 0$ under your assumptions.

Expand the determinant $D=\det k(x_j,y_k)$ along the first row. This shows that as a function of $x=x_1$, it is of the form $$ D(x) = \sum_{j=1}^n \frac{c_j}{1-y_j x} , $$ with $c_j$ independent of $x=x_1$ (and of $y_j$, but that doesn't matter here).

This rational function has $n$ poles at $1/y_1,\ldots, 1/y_n$. On the other hand, clearly $D(x_2)=\ldots =D(x_n)=D(\infty)=0$, and since this is a total of $n$ zeros, we have found all of them. Thus $D\not= 0$ under your assumptions.

Small detail added later: If $y_j=0$ for some $j$, then we need to slightly modify the argument: now we have $n-1$ poles, and also $n-1$ zeros since now $D(\infty)=c_j\not= 0$. (This last fact we could also have obtained directly from an inductive argument since the $c_j$ are determinants of $(n-1)\times (n-1)$ submatrices of the same type.)