The vertices of a triangle are $(pq,\frac{1}{pq}),(qr,\frac{1}{qr})$ and $(rq,\frac{1}{rp})$ where p, q, r are the roots of the equation$y^3-3y^2+6y+1=0$ . The coordinates of its centroid are

p, q, r, are the roots of equation $y^3-3y^2+6y+1=0$ 

so, $p+q+r=3,pq+qr+rp=6$ and $pqr=-1$ , now the centroid of the triangle is   $\left(\frac{pq+qr+rp}{3},\frac{\frac{1}{pq}+\frac{1}{qr}+\frac{1}{rp}}{3}\right)$ i.e.

 = $\left(\frac{pq+qr+rp}{3},\frac{p+q+r}{3pqr}\right)=\left(\frac{6}{3},\frac{3}{-3}\right)$ or $\left(2,-1\right)$