Let's look at what happens when derivative of the determinant of an invertible matrix $A$ with respect to itself. Apparently, the result is a matrix, say $D$, then we have
$$D^q_p=\left(\frac{\partial\det A}{\partial A}\right)^q_p=\frac{\partial(\epsilon_{i_1\ldots i_n}a^{i_1}_1\cdots a^{i_n}_n)}{\partial a^p_q}$$
where $\epsilon$ is Levi-Civita symbol and Einstein notation is used. If $A$ has no special structure that each of $n^2$ components are seen as independent, we can proceed to
$$D^q_p=\epsilon_{i_1\ldots i_n}a_1^{i_1}\cdots\frac{\partial a^{i_p}_q}{\partial a^p_q}\cdots a_n^{i_n}$$
However when $A=A^T$ is symmetric, each product has 2 terms that depend on $a^p_q$, hence
$$D^q_p=\epsilon_{i_1\ldots i_n}\left(a_1^{i_1}\cdots\frac{\partial a^{i_p}_p}{\partial a^p_q}\cdots a^{i_q}_q\cdots a_n^{i_n}+a_1^{i_1}\cdots a^{i_p}_p\cdots\frac{\partial a^{i_q}_q}{\partial a^p_q}\cdots a_n^{i_n}\right)$$
This leads to different forms of derivative. Yet it is now clear that the derivative of an invertible matrix w.r.t a scalar doesn't depend on its structure.
For typing convenience, let me substitute Latin in place of your Greek letters
$$\eqalign{
X &= \Xi \cr
x &= \vartheta = \operatorname{vec}(X) \cr
W &= \Omega = I + XX^T = W^T \cr\cr
}$$
Then for your first function, the differential and gradient can be calculated as
$$\eqalign{
f &= \log\det W \cr
\cr
df &= d\log\det W = d\operatorname{tr}\log W \cr
&= W^{-T}:dW \cr
&= W^{-T}:(dX\,X^T+X\,dX^T) \cr
&= \big(W^{-T}X+W^{-1}X\big):dX \cr
&= 2\,W^{-1}X:dX \cr
&= 2\,\operatorname{vec}(W^{-1}X)\cdot\,dx \cr
\cr
\frac{\partial f}{\partial x} &= 2\,\operatorname{vec}(W^{-1}X) \cr\cr
}$$
In your second function, I don't quite understand the definition of $\,H(\vartheta)\,$ so all I can offer is a partial solution
$$\eqalign{
M &= W^{-1} = M^T \cr
f &= H:MH \cr
\cr
df &= dH:MH + H:M\,dH + H:dM\,H \cr
&= (M+M^T)H:dH + HH^T:dM \cr
&= 2\,MH:dH - HH^T:M\,dW\,M \cr
&= 2\,MH:dH + MHH^TM:dW \cr
&= 2\,MH:dH + MHH^TM:(dX\,X^T+X\,dX^T) \cr
&= 2\,MH:dH + 2\,MHH^TMX:dX \cr
\cr
}$$
You can finish off the solution by expanding $dH$ in terms of $dX$, then vectorizing.
In the above, a colon denotes the double-dot (aka Frobenius) product, which is merely a product notation for the trace, i.e. $$A:B=\operatorname{tr}(A^TB)$$
Update
I'm guessing that $H$ is a partitioned matrix: $H=[\,X, I\,]$
Expanding that term in the differential
$$\eqalign{
MH:dH &= [\,MX, M\,]:[\,dX, 0\,] \cr
&= MX:dX + M:0 \cr
}$$
So, continuing with the full differential
$$\eqalign{
df &= 2\,\Big(MX + MHH^TMX\Big):dX \cr
&= 2\,\operatorname{vec}(MX + MHH^TMX)\cdot dx \cr
\cr
\frac{\partial f}{\partial x} &= 2\,\operatorname{vec}(MX + MHH^TMX) \cr
}$$
Best Answer
Hint: use $(X^{-1})_{ij}=\frac{C_{ji}}{\operatorname{det}(X)}$, with $C_{ji}=(-1)^{i+j}X_{ji}=(-1)^{i+j}X_{ij}$, as $X$ is symmetric.
Then
$$\frac{\partial (X^{-1})_{ij}}{\partial X_{kl}}= \frac{\partial }{\partial X_{kl}}\left( \frac{(-1)^{i+j}X_{ij}}{\operatorname{det}(X)}\right).$$
Using the formula for the derivative of the determinant of $X$
$$\frac{\partial \det(X)}{\partial X_{kl}}= \det(X)(X^{-1})_{lk}$$
you can arrive at the result.