If you're trying to do the calculus on a vector space, remember that the derivative of
$$f:\Bbb R^n\to\Bbb R^m$$
is an $m\times n$ matrix. In particular, if $m\ne n$ how do you plan to talk about semi-definiteness? And even if you have $m=n$, recall that semi-definite matrices are also symmetric. The partial derivatives at a point may not be equal everywhere, so what if ${\partial^2f\over\partial y\partial x} \ne{\partial^2f\over\partial x\partial y}$?
Since you want the more special case of $f:\Bbb R^n\to \Bbb R$ where a notion of "increasing" is a little more specific, then there is the tool of the directional derivative which can tell you about increasing or decreasing in a specific direction, but again this isn't really related to the semi-definiteness. Especially of note is that the derivative is a vector, not a number, and definitely not a (symmetric) $n\times n$ matrix of any sort. Note that in this case if you are increasing in one direction you are decreasing in the opposite direction, so it's hard to make a canonical choice for which "increasing" is clearly what you mean. Unlike maps $g:\Bbb R\to\Bbb R$ where you have a notion of increasing in the domain, maps from spaces without an ordering are harder to pin down. If you can produce a total ordering on your domain, you could make it more precise, but it's not clear what you do when two things are only in a partial order.
In particular: it's not clear any simple fact about the derivative will correspond to the monotonicity in an arbitrary order, which I think is what you're looking for.
In your example: I don't think you want the $X$ matrices in there, you want just two matrices, $X\le Y$ where $\le $ means the difference is positive semi-definite. So we want to show $\log |X|\le \log |Y|$. Then
$$\log |Y|-\log |X|=\log |YX^{-1}|.$$
Two things to note: ($1$) if you don't demand definite rather than semi-definite, you can have a zero matrix, which makes the log thing undefined. So this is no good unless you demand definite. Also, you can only define this on things which are already positive determinant, since a negative determinant gives $\log$ of a negative number, which is undefined. But then $|XY^{-1}|$ has a negative log iff $|XY^{-1}|<1\iff 0<|X|\le |Y|$.
So if we find $X,Y$ with $|X|<|Y|$ but $Y-X$ is positive definite, we have a contradiction. For this choose
$$X=\begin{pmatrix} 1 & 0 \\ 0 & 1\end{pmatrix}, Y=\begin{pmatrix} 2 & -6 \\ 1 & 2\end{pmatrix}$$
then
$$Y-X = \begin{pmatrix} 1 & -6 \\ 1 & 1\end{pmatrix}, |X|=1, |Y|=2.$$
The sub-determinants of $Y-X$ are $1, 7$ respectively, so it is positive definite. So this is a counterexample for that ordering.
At the same time I think your example misses the forest for the trees.
The point is that there's not a simple test of when something is increasing based on the derivatives. Whatever it would mean to be increasing at a matrix, there is not going to be a property of a given matrix which relates to the derivative in a tidy way to talk about increasing. In particular, you want, psychologically, for the positive definiteness property of a matrix to be analogous to the positive quality of a real number and for this to translate into an easy test on matrix functions for monotonicity.
The problem If you look at the original context you got this from the sign of the derivative takes place in the domain, not the domain. Your question asks about the domain, which is a totally different animal. Now, you might alter the question to talk about the derivative, which is a vector, but--as I have said earlier--there is no canonical notion of positivity for a vector, so there's no way to phrase your question in a way which is actually analogous to the original test you are motivated by. You've imposed some really complicated conditions, but none of them actually have anything to do with your original question, they're just bending it into a direction which is unrelated to the original, and which tries to talk about monotonicity in a totally derivative-unrelated way.
Best Answer
As far as I can tell, the only thing you can do is use a current candidate for the minimum, say $y$, and then guarantee that the global minimum, if it is not $y$, must occur in the region $f(x) < y$ and $g(x) < y$ since your functions are positive. So you can restrict the region of your search.