Consider a real positive definite matrix $A\in R^{d\times d}$. Let $S_d^+$ be the set of symmetric positive matrices. Is it always true that $$(x,B) \to x^TB^{-1/2} A B^{-1/2} x$$ is jointly convex in $(x,B)\in R^d \times S_d^+$?
It is true in the case $A=I$, see Proving that $(x,Q) \mapsto \langle x,Q^{-1}x\rangle$ is convex for instance for a proof.
It is not even convex in $B$ alone: $$ A = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} ,\quad B_1 = \begin{pmatrix} 1 & 0.1 \\ 0.1 & 0.02 \end{pmatrix} ,\quad B_2 = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} ,\quad x_1 = \begin{pmatrix} 1 \\ 0.5\end{pmatrix} .$$
Plotting
$$
t \mapsto f(x, t B_1 +(1-t) B_2)
$$
for $t \in [0,1]$
yields a non-convex graph: 
