The question does not appear to have anything to do with $\Omega$ or with derivatives, since you are asking for a pointwise bound. It could be written as $\|A\|\le (C(\epsilon)+\epsilon \|A\|) \operatorname{tr} A^{-1}$.
Assume $A$ diagonal, as we can do because diagonalization preserves $\operatorname{tr} A^{-1}$. Let $\lambda_1\ge \dots \ge \lambda_n>0$ be the diagonal entries. Then $\operatorname{tr} A^{-1}\ge \lambda_n^{-1}$. Since $\mu>\lambda_1\cdots\lambda_n\ge \lambda_1\lambda_n^{n-1}$, it follows that $\lambda_n\le \lambda_1^{-1/(n-1)}\mu^{1/(n-1)}$. Hence $\operatorname{tr} A^{-1}\ge \lambda_1^{1/(n-1)}\mu^{-1/(n-1)}$. Let's simplify notation by writing $\operatorname{tr} A^{-1}\gtrsim \lambda_1^{1/(n-1)}$ where $\gtrsim$ indicates the presence of a multiplicative constant that may involve $\mu$ and $n$.
The matrix norm $\|A\|$ (whatever it is) is comparable to $\lambda_1$ in the sense that $\lambda_1 \lesssim \|A\|\lesssim \lambda_1$. Therefore, our task reduces to showing that $\lambda_1 \lesssim C(\epsilon)\, \lambda_1^{1/(n-1)} + \epsilon\, \lambda_1^{n/(n-1)}$ for all $\lambda_1>0$. But this is straightforward: no matter how small $\epsilon>0$, we have $\lambda_1 \le \epsilon\, \lambda_1^{n/(n-1)}$ for all sufficiently large $\lambda_1$. Then choose $C(\epsilon)$ so that $\lambda_1\le C(\epsilon) \, \lambda_1^{1/(n-1)}$ for all $\lambda_1$ that are not sufficiently large.