jmc

Observe that $$\begin{array}{rl}p(0,0)& =a_0=0\\ p(1,0)& =a_0+a_1+a_3+a_6=a_1+a_3+a_6=0\\ p(-1,0)& =-a_1+a_3-a_6=0.\end{array}$$Adding the above two equations gives $a_3=0$, and so we can deduce that $a_6=-a_1$. Similarly, plugging in $(0,1)$ and $(0,-1)$ gives $a_5=0$ and $a_9=-a_2$. Now, $$\begin{array}{rl}p(1,1)& =a_0+a_1+a_2+a_3+a_4+a_5+a_6+a_7+a_8+a_9\\ & =0+a_1+a_2+0+a_4+0-a_1+a_7+a_8-a_2=a_4+a_7+a_8=0\\ p(1,-1)& =a_0+a_1-a_2+0-a_4+0-a_1-a_7+a_8+a_2\\ & =-a_4-a_7+a_8=0\end{array}$$Therefore, $a_8=0$ and $a_7=-a_4$. Finally, $$p(2,2)=0+2a_1+2a_2+0+4a_4+0-8a_1-8a_4+0-8a_2=-6a_1-6a_2-4a_4=0.$$Hence, $3a_1+3a_2+2a_4=0$. Now, $$\begin{array}{rl}p(x,y)& =0+a_{1}x+a_{2}y+0+a_{4}xy+0-a_1{x}^3-a_4{x}^{2}y+0-a_2{y}^3\\ & =a_{1}x(1-x)(1+x)+a_{2}y(1-y)(1+y)+xy(1-x)a_4\\ & =a_{1}x(1-x)(1+x)+a_{2}y(1-y)(1+y)-\left(\frac{3}{2}{a}_1+\frac{3}{2}{a}_2\right)xy(1-x)\\ & =a_1\left(x-x^3-\frac{3}{2}xy(1-x)\right)+a_2\left(y-y^3-\frac{3}{2}xy(1-x)\right).\end{array}$$If $p(r,s)=0$ for every such polynomial, then $$\begin{array}{rl}r-r^3-\frac{3}{2}rs(1-r)& =0,\\ s-s^3-\frac{3}{2}rs(1-r)& =0.\end{array}$$These factor as $$\begin{array}{rl}\frac{1}{2}r(1-r)(2r-3s+2)& =0,\\ \frac{1}{2}s(3r^2-3r-2s^2+2)& =0.\end{array}$$Hence, $r=0,$ $r=1,$ or $r=\frac{3s-2}{2}.$

Substituting $r=0$ into the second equation, we get $s^3=s,$ so $s=-1,$ 0, or 1.

Substituting $r=1$ into the second equation, we again get $s^3=s,$ so $s=-1,$ 0, or 1.

Substituting $r=\frac{3s-2}{2}$ into the second equation, we get $$s-s^3-\frac{3}{2}\cdot \frac{3s-2}{2}\cdot s\cdot \left(1-\frac{3s-2}{2}\right)=0.$$This simplifies to $19s^3-54s^2+32s=0,$ which factors as $s(s-2)(19s-16)=0.$ We are looking for a value where $s$ is not an integer, so $s=\frac{16}{19}.$ Then $r=\frac{5}{19},$ so $(r,s)=\boxed{\left(\frac{5}{19},\frac{16}{19}\right)}.$

This is an instance of a result known as Bezout's Theorem, from algebraic geometry. Loosely speaking, Bezout's Theorem states that if we plot two curves, then the number of intersection points is equal to the product of their degrees. Here, one curve is $$x(x-1)(2x-3y+2)=0,$$shown in red below, which consists of three lines. The other curve is $$y(3x^2-3x-2y^2+2)=0,$$shown in blue below, which consists of a line and a hyperbola. The degree of both curves is 3. Note how the red and blue curves intersect at the eight given points, so by Bezout's Theorem, there is a ninth point of intersection, which is exactly $\left(\frac{5}{19},\frac{16}{19}\right).$