Answer
$(-\infty,-2)\cup[-1/3,1)\cup(3,\infty)$
Work Step by Step
1. Move all terms to one side $\frac{1}{x-3}+\frac{1}{x+2}-\frac{2x}{x^2+x-2}\geq0$
$\frac{(x+2)(x-1)+(x-3)(x-1)-2x(x-3)}{(x-3)(x+2)(x-1)}
=\frac{(x+2)(x-1)-(x-3)(x+1)}{(x-3)(x+2)(x-1)}$
$f(x)=\frac{(3x+1)}{(x-3)(x+2)(x-1)}\geq0$
2. Cut points $-2,-1/3,1,3$, build up a sign table as shown
3. Solution $(-\infty,-2)\cup[-1/3,1)\cup(3,\infty)$
