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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. 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.