Doing this substitution, the equation becomes $$y - \dfrac{3}{y} = 2,$$ which is equivalent to $$y^2 -2y-3=0.$$ This has solutions $y = 3$, and $y = -1$. So we have $$\dfrac{3x-1}{x^2+1} = 3, $$ which has no real solutions, or $$\dfrac{3x-1}{x^2+1} = -1, $$ which has solutions $x = 0$ and $x = -3$.