This is a problem from Freidberg linear algebra (4th edition) chapter 6.4.21
The whole problem and the hints are
Let $V$ be a finite-dimensional inner product space, and let $T$ and $U$ be self-adjoint operators on $V$ such that $T$ is positive definite. Prove that both $TU$ and $UT$ are diagonalizable linear operators that have only real eigenvalues. Hint: Show that $UT$ is self-adjoint with respect to the inner product $<x, y>' = <T(x), y>$ To show that $TU$ is self-adjoint, repeat the argument with the inverse of $T$ in place of $T$.
I have tried
$<TU(x), y> = <U(x), T^*(y)> = <U(x), T(y)> = < x, U^*T(y) > = <x, UT(y)>$
But this didn't work.
I don't know why the hint is given, and I think there is another way of not using that hint.
Because I think "Show that $UT$ is self-adjoint with respect to the inner product $<x,y>'=<T(x),y>$" is a bit contrived.