The update for Davidon-Fletcher-Powell (DFP) is given as the following:
$$ B_{k+1} = (I - \rho_ky_ks_k^{\top})B_k(I - \rho_ks_ky_k^{\top})+\rho_ky_ky_k^{\top} $$
where $y_k,s_k \in \mathbb{R}^n$ such that $\rho_k=y_k^{\top}s_k>0$ and $B_ks_k=y_k$.
Show that when $B_k$ is positive definite so is $B_{k+1}$.
Let $x \in \mathbb{R}^n$, then
$$x^{\top} B_{k+1}x = x^{\top}(I - \rho_ky_ks_k^{\top})B_k(I - \rho_ks_ky_k^{\top})x+\rho_kx^{\top}y_ky_k^{\top}x.$$
As $B_k$ is positive definite, we have
$$ x^{\top}(I - \rho_ky_ks_k^{\top})B_k(I - \rho_ks_ky_k^{\top})x \geq 0.$$
Moreover,
$$ \rho_k x^{\top}y_ky_k^{\top}x \geq 0.$$
Hence, $$x^{\top} B_{k+1}x \geq 0.$$
To show that $B_{k+1}$ is positive definite, it remains to show that the inequality is strict when $x \neq 0$, or equivalently, $x^{\top} B_{k+1}x = 0$ implies $x=0$. From above $x^{\top} B_{k+1}x = 0$ implies $$(I - \rho_ks_ky_k^{\top})x = 0,$$ and $$y_k^{\top}x =0.$$
But this implies $x=0$. Hence, $B_{k+1}$ is positive definite.