Let's say I have an ellipsoid centered at $x^*$: $${\cal E} = \{ y \in \mathbb R^n : (y - x^*)^T A (y - x^*) \leq R^2 \}$$ where $A$ here is an invertible $n \times n$ symmetric matrix.
Now, let's say I have another symmetric matrix $E$ with $$- 3 A \preccurlyeq E \preccurlyeq 3 A$$ (here, $- 3 A \preccurlyeq E$ means $E + 3 A$ is a psd and the same for other).
Then, I want to show that for any $y$ in the ellipsoid,
$$(y - x^*)^T E A^{-1} E (y - x^*) \leq 81 R^4$$
To do so, I was thinking I can show
$$ - N \preccurlyeq A \preccurlyeq N, - M \preccurlyeq B \preccurlyeq M \implies A B \preccurlyeq M N$$ in general, and then using this to argue
$$E A^{-1} E \preccurlyeq 9 R^2 A \implies (y - x^*)^T E A^{-1} E (y - x^*) \\ \leq (y - x^*)^T 9 R^2 A (y - x^*) \leq 9 R^4 \leq 81 R^4$$,
but I am not sure if what I am trying to prove is even correct.
Here is a partial answer if $A$ and $E$ are both PSD. In general, the following holds:
$$ A \succeq B \implies \mathsf{tr}(AX) \geq \mathsf{tr}(BX), $$ where $X \in \mathbb{S}_{+}^n$ is an arbitrary PSD matrix.
Now let $z \in \mathbb{R}^n$. Since $E \preceq 3A$, it follows that $A^{-1} \preceq 3 E^{-1}$. Consequently, \begin{align} z^{\mathsf{T}} E A^{-1} E z = \mathsf{tr}(A^{-1} (Ez) (Ez)^{\mathsf{T}}) &\leq 3 \mathsf{tr}(E^{-1} (Ez) (Ez)^{\mathsf{T}}) \\ &\leq 3 \mathsf{tr}(z^{\mathsf{T}} E E^{-1} E z) \\ &= 3 z^{\mathsf{T}} E z \\ &\leq 9 z^{\mathsf{T}} A z. \end{align}