Closed-form solution for the determinant of a Vandermonde-like matrix

These seem to be a direct consequence of Schendel's 1891 theorem about "confluent Vandermonde matrices", as explained in (say) "On a Recursion Formula Related to Confluent Vandermonde", Shui-Hung Hou and Edwin Hou, The American Mathematical Monthly, Vol. 122, No. 8 (October 2015), pp. 766-772, or here or here. An example of a confluent Vandermonde determinant is $$ \begin{vmatrix}x^3&3x^2&y^3&3y^2\\x^2&2x&y^2&2y\\x&1&y&1\\1&0&1&0\end{vmatrix}$$ where certain columns of an ordinary Vandermonde determinant have been replaced by derivatives of others. Schendel's formula is that this determinant is equal to $(x-y)^4$. If you subtract the third column from the first, and then expand by minors, you end up with the identities $$ \begin{vmatrix}x^3&3x^2&y^3&3y^2\\x^2&2x&y^2&2y\\x&1&y&1\\1&0&1&0\end{vmatrix} = \begin{vmatrix}x^3-y^3&3x^2&y^3&3y^2\\x^2-y^2&2x&y^2&2y\\x-y&1&y&1\\0&0&1&0\end{vmatrix} = -\begin{vmatrix}x^3-y^3&3x^2&3y^2\\x^2-y^2&2x&2y\\x-y&1&1\end{vmatrix}. $$ This is the OPs $n=3$ example.

Schendel's general formula for confluent Vandermonde determinants is $\prod_{i<j}(x_i-x_j)^{n_in_j}$, where the value $x_i$ is used to form $n_i$ columns consisting of the first $n_i$ derivatives (the $0$-th up through the $n_i-1$-th derivative) of the usual Vandermonde column $(1,x_i, x_i^2,\ldots)'$. In the OP's case, all of the $n_i=2$. The ordinary Vdm formula has all $n_i=1$.