The derivative can be found via implicit differentiation. That is, 
$$
\frac{\mathrm{d}\operatorname{vec}\left(Y\right)}{\mathrm{d}\operatorname{vec}\left(X\right)} = \left(\frac{\mathrm{d} \operatorname{vec}\left(X\right)}{\mathrm{d}\operatorname{vec}\left(Y\right)}\right)^{-1}.$$
It is relatively easy to compute the derivative of $A$ with respect to $f(A)$ since $A = f(A)f(A)^{\top}$. The only trick part is restricting $f(A)$ to be lower triangular.

For general $X$, we have
$$
\frac{\mathrm{d} \operatorname{vec}\left(XX^{\top}\right)}{\mathrm{d} \operatorname{vec}\left(X\right)} = \left(I + K\right)\left(X\otimes I\right),$$
where $K$ is the [Commutation Matrix][1].

Now to get the derivative with respect to the $\operatorname{vech}$ requires use of the chain rule. This gives
$$
\frac{\mathrm{d} \operatorname{vech}\left(XX^{\top}\right)}{\mathrm{d} \operatorname{vech}_{\Delta}\left(X\right)} = L \left(I + K\right)\left(X\otimes I\right) D,$$
where here $L$ is the elimination matrix, and $D$ is the "lower triangular duplication matrix" which has the property that $D \operatorname{vech}\left(M\right) = \operatorname{vec}\left(M\right)$ for lower triangular matrices $M$. The sought derivative is the matrix inverse of the above expression. 


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