Answer
See graphs, $\left(1,3\right),\left(3,1\right),\left(-3,-1\right),\left(-1,-3\right)$
Work Step by Step
1. See graphs for $x^2+y^2=10$ and $xy=3$
2. The 2nd equation gives $y=\frac{3}{x}$, plug-in the 1st equation to get $x^2+\frac{9}{x^2}=10$ or $x^4-10x^2+9=0$ or $(x^2-1)(x^2-9)=0$, thus $x=\pm1, \pm3$
3. Use the 2nd equation with the above results, we can find the intersect(s): $\left(1,3\right),\left(3,1\right),\left(-3,-1\right),\left(-1,-3\right)$
