Answer
$(3,1)$, $(-3,-1)$, $(1,3)$, $(-1,-3)$
Work Step by Step
Here, we have $y=\dfrac{3}{x}$
Now, $x^2+(\dfrac{3}{x})^2=10$
This gives: $x^4-10x^2+9=0$
This yields : $(x^2-9) (x^2-1)=0$
Thus, we have $x=3,-3, 1,-1$
when $x=3$ then we have $y=\dfrac{3}{3}=1$
when $x=-3$ then we have $y=\dfrac{3}{-3}=-1$
when $x=1$ then we have $y=\dfrac{3}{1}=3$
when $x=-1$ then we have $y=\dfrac{3}{-1}=-3$
Hence, our answers are: $(3,1)$, $(-3,-1)$, $(1,3)$, $(-1,-3)$
