I have that
$$ y = (x'Bx)^{-1}x'A$$
where $x$ is $k \times r$, $A$ is $k \times 1$, and $B$ is $k \times k$ and invertible.
What is the expression for the Jacobian of $y$ with respect to $x$?
I have so far that it is
$$ (x'Bx)^{-1}A'+(x'Bx)^{-1}(2x'B)(x'Bx)^{-1}x'A $$
but this doesn't feel quite right to me. Can someone point me in the right direction?
Normally a Jacobian is a matrix whose columns are partial derivatives. Other than this picky note, the tools you need to solve this should all be on The Matrix Cookbook, unless you need some sort of elegant solution.
In particular, what you've made so far seems correct for vector $x$, but for matrix $x$, I'm unsure if you are right and even if you are I suspect you skipped some steps. See for instance: