$-\frac{1}{x^2+x-6}\le \frac{1}{1-x^2}$

$\frac{1}{1-x^2}+\frac{1}{x^2+x-6}\ge 0\Rightarrow \frac{1}{\left(1+x\right)\left(1-x\right)}+\frac{1}{\left(x-2\right)\left(x+3\right)}\ge 0\Rightarrow \frac{x^2+x-6+1-x^2}{\left(1+x\right)\left(1-x\right)\left(x-2\right)\left(x+3\right)}\ge 0\Rightarrow$

$\Rightarrow \frac{x-5}{\left(1+x\right)\left(1-x\right)\left(x-2\right)\left(x+3\right)}\ge 0\Rightarrow x\in ]-\infty ,-3[\cup ]-1,1[\cup ]2,5]$