Can someone explain how to solve linear inequality? 
Can someone explain how to solve this linear inequality?

Multiply the both sides by $x^2-9=(x-3)(x+3)$. 
But note that you need to have two cases as $x^2-9\gt 0$ or $x^2-9\lt 0$.
(Case 1)
If $x^2-9\gt 0\iff x\lt -3\ \text{or}\ x\gt 3$, then 
$$\frac{6(x^2-9)}{x^2-9}+\frac{(13-x)(x+3)(x-3)}{x+3}\le\frac{3(x+3)(x-3)}{x+3}-\frac{2(x+3)(x-3)}{-(x-3)}$$
i.e.
$$6+(13-x)(x-3)\le 3(x-3)+2(x+3)\iff x\le 5\ \text{or}\ x\ge 6.$$
Hence, in this case, we have
$$x\lt -3\ \ \text{or}\ \ 3\lt x\le 5\ \ \text{or}\ \ x\ge 6.$$
(Case 2)
If $x^2-9\lt 0\iff -3\lt x\lt 3$, then 
$$\frac{6(x^2-9)}{x^2-9}+\frac{(13-x)(x+3)(x-3)}{x+3}\color{red}{\ge}\frac{3(x+3)(x-3)}{x+3}-\frac{2(x+3)(x-3)}{-(x-3)}$$
i.e.
$$6+(13-x)(x-3)\ge 3(x-3)+2(x+3)\iff 5\le x\le 6.$$
Hence, in this case, there is no such $x\in\mathbb R$.
Hence, the answer is
$$x\lt -3\ \ \text{or}\ \ 3\lt x\le 5\ \ \text{or}\ \ x\ge 6.$$
