It occurred to me to compute the derivative of the inverse instead, which should be simple. Using the chain rule, I first convinced myself that
$$
\frac{\mathrm{d} \operatorname{vec}\left(X X^{\top}\right)}{\mathrm{d} \operatorname{vec}\left(X\right)} = \left(I + C\right)\left(I \otimes X\right),
$$
where $C$ is the [Commutation Matrix][1]. From there I computed the inverse as
$$
\frac{\mathrm{d} \operatorname{vec}\left(X\right)}{\mathrm{d} \operatorname{vec}\left(X X^{\top}\right)} = \left[ 
\left(\frac{\mathrm{d} \operatorname{vec}\left(X X^{\top}\right)}{\mathrm{d} \operatorname{vec}\left(X\right)}\right)^{\top}
 \right]^{-1},
$$
where the power is element-wise or 'Hadamard' power. Wrapping this with $L$, the elimination matrix, gives the desired result.



  [1]: https://en.wikipedia.org/wiki/Commutation_matrix "commutation matrix"