Let $D\in\mathbb{R}^{n\times n}$ be a positive definite diagonal matrix and $A\in\mathbb{R}^{n\times n}$ be a symmetric matrix. We know $DA$ has real eigenvalues because $DA=D^{0.5}(D^{0.5}AD^{0.5})D^{-0.5}$. Therefore, $DA$ is similar to $B:=D^{0.5}AD^{0.5}$, which is a symmetric matrix.
I was wondering if the eigenvectors of matrix $DA$ are mutually orthogonal (as this is the case for a symmetric matrix)?
$DA$ has not necessarily mutually orthogonal eigenvectors. Take for example
$$D= \begin{pmatrix} 2 & 0 \\ 0 & 1 \end{pmatrix}, \qquad A = \begin{pmatrix} 1 & -1 \\ -1 & 1 \end{pmatrix}.$$
Then
$$DA = \begin{pmatrix} 2 & -2 \\ -1 & 1 \end{pmatrix}.$$
The eigenvalues of $DA$ are $0$ and $3$ and two corresponding eigenvectors are $v = (1, 1)$ and $w = (-2, 1)$, so $v$ and $w$ are not orthogonal.
Note that the statement can be easily proven to be true if, for example, $D$ and $A$ commute.