Prove: A triangle inscribed in a rectangular hyperbola has its orthocenter on that hyperbola

Let $A = (p,1/p), ~B = (q,1/q),~ C = (r, 1/r)$ be the three points of the triangle, let $H$ be the orthocenter.

Verify that the slope of $AB$ is exactly $-1/pq$, so the slope of the altitude $h_C$ is $pq$. Similarly, the slope of the altitude $h_A$ is $qr$.

Now you can calculate the position of $H$, by intersecting $h_A$ and $h_C$: We have $$A + x\cdot\begin{pmatrix}1\\qr\end{pmatrix} = H = C + y \cdot \begin{pmatrix}1\\pq\end{pmatrix}$$

for some $x$ and $y$. You can solve this linear equation system for $x$ and $y$ and get that $H = (-1/pqr, -pqr)$. This point clearly lies on your given hyperbola.

Let we start with a straightforward

Lemma 1. If $A,B$ are two points on a rectangular hyperbola and $ACBD$ is a rectangle with its sides being parallel to the hyperbola axis, the centre $O$ of the hyperbola lies on the $CD$ line.

Such almost obvious fact has an important consequence: let $ABC$ a triangle with its vertices lying on a rectangular hyperbola and let $M_A,M_B,M_C$ be the midpoints of $BC,AC,AB$. The centre $O$ of our hyperbola fulfills $$ \widehat{M_A O M_C} = \widehat{A B C} $$ by the previous lemma, hence

Lemma 2. If $A,B,C$ are three distinct points on a rectangular hyperbola with centre $O$,
$O$ belongs to the nine point circle of $ABC$.

If $H$ is the orthocenter of $ABC$, the midpoints of $HA,HB,HC$ also lie on the nine point circle of $ABC$. It follows that the nine-point circles of $ABC,HAB,HAC,HBC$ are the same circle.

Given three distinct points $A,B,C$ in the plane and two perpendicular directions, by Lemma 1 there is only one rectangular hyperbola through $A,B,C$ with axis parallel to the given directions.
Moreover,

Lemma 3. Given a hyperbola (or ellipse) and two parallel chords $AB,CD$ on it, the centre of the hyperbola (ellipse) lies on the segment joining the midpoint of $AB$ with the midpoint of $CD$. The same apply if we replace a chord with a tangent at some point.

Now it is not difficult to figure that our lemmata proves that the hyperbola fixed by $H,A,B$ is the same as the hyperbola fixed by $A,B,C$, so $H$ lies on the hyperbola fixed by $A,B,C$, as wanted.

We may also exploit the fact that any circumconic through the orthocenter of a triangle is a rectangular hyperbola (see Jerabek's hyperbola).