$3=\dfrac {\sqrt{x+1}+\sqrt{x-1}}{\sqrt{x+1}-\sqrt{x-1}}=$

$\dfrac {\sqrt{x+1}+\sqrt{x-1}}{\sqrt{x+1}-\sqrt{x-1}}\dfrac{\sqrt{x+1}+\sqrt{x-1}}{\sqrt{x+1}+\sqrt{x-1}}=$

$\dfrac{(x+1)+2\sqrt{x^2-1}+(x-1)}{(x+1)-(x-1)}=$

$\dfrac {2x+2\sqrt{x^2-1}}{2}=3$

$x+\sqrt{x^2-1}=3$

$\sqrt{x^2-1}=3-x$

$x^2-1=9-6x+x^2$

$6x=10$

$x=\dfrac 5 3$

This is the only real solution.