If $A$ and $B$ are symmetric positive definite then $\vec{x}^TA\vec{x}>0$ and $\vec{x}^TB\vec{x}>0$ as well as $A^T=A$ and $B^T=B$. Here's where the confusion lies, when showing product of $A\cdot B$ is symmetric positive definite. I tried $(A\cdot B)^T=B^T\cdot A^T=B\cdot A=A\cdot B$, but the last equality is false, consider counterexample $A=\begin{bmatrix}1&1&1\\ 1&1&1\\ 1&1&1\end{bmatrix}$ and $B=\begin{bmatrix}1&0&1\\ 0&1&0\\ 1&0&1\end{bmatrix}$.
I also need help in showing $A\cdot B$ positive definite, does it immediately follow vacuously?
Edit: Assume both $A$ and $B$ are diagonalizable so that commutativity is allowed.
For showing product is positive definite, I'm doing this but now I'm stuck. $\vec{x}^TAB\vec{x}=\vec{x}^TA^TB\vec{x}=(A\vec{x})^T(B\vec{x})=\vec{a}^TI\vec{b}>0$ where $B\vec{x}=\vec{b}$ and $A\vec{x}=\vec{a}$ and $I_{n\times n}$ is the identity matrix and it is positive definite
$A=\begin{bmatrix}1&0\\0&2\end{bmatrix},$ $B=\begin{bmatrix}2&1\\1&2\end{bmatrix}.$ $AB=\begin{bmatrix} 2 &1\\2&4 \end{bmatrix},$ $BA=(AB)^T=\begin{bmatrix}2&2\\1&4\end{bmatrix},$ so "both matrices are diagonalizable$\Rightarrow$commutativity" is false. As @Scounged point out, simultaneously diagonalizable can. Because $$AB=(PD_AP^{-1}) (PD_BP^{-1})=P D_B D_A P^{-1}=(PD_BP^{-1})(PD_AP^{-1}) =BA.$$ As diagonal matrixes are commutative.
So $AB$ may not be symmetric. Also, it may not be positive definite. Like $A=\begin{bmatrix}2&-4\\-4&10\end{bmatrix},$ $B=\begin{bmatrix}5&-4\\-4&4\end{bmatrix}.$ $AB=\begin{bmatrix} 26 &-24\\-60&56 \end{bmatrix}.$ (Just searched on Internet, many examples.)
However, if you suppose $AB=BA,$ then they can be simultaneously diagonalized, and the eigenvalue of $AB$ is just $A$'s times $B$'s, so it's positive definite. Also symmetric, of course.