Step 1: Find slopes of the given pair of lines.
Setting $y=mx$ in $y^2-8xy-9x^2=0$: $m^2-8m-9=0 \implies (m-9)(m+1)=0$. Slopes: $m_1=9$ and $m_2=-1$.
Step 2: Identify the slopes of the triangle sides.
The side bisected perpendicularly by $y=9x$ (slope 9) has slope $-1/9$; the side bisected by $y=-x$ (slope $-1$) has slope $1$.
Step 3: Reflect $A=(\alpha,\beta)$ in $y=9x$.
Reflection formula for line $y=mx$: for $m=9$, $1+m^2=82$. \[B=\left(\frac{(1-81)\alpha+18\beta}{82},\;\frac{18\alpha+(81-1)\beta}{82}\right)=\left(\frac{-80\alpha+18\beta}{82},\;\frac{18\alpha+80\beta}{82}\right).\]
Step 4: Reflect $A=(\alpha,\beta)$ in $y=-x$.
For $m=-1$, $1+m^2=2$: $C=(-\beta,\,-\alpha)$.
Step 5: Compute centroid $G$ of triangle $A,B,C$.
\[G_x=\frac{\alpha+(-80\alpha+18\beta)/82+(-\beta)}{3}=\frac{(2\alpha-64\beta)/82}{3}=\frac{\alpha-32\beta}{123}.\] \[G_y=\frac{\beta+(18\alpha+80\beta)/82+(-\alpha)}{3}=\frac{(-64\alpha+162\beta)/82}{3}=\frac{81\beta-32\alpha}{123}.\]
Step 6: State the centroid.
\[ \boxed{\frac{1}{123}(\alpha-32\beta,\;81\beta-32\alpha)} \]