Find $a,b$ at $f(x)=\frac{x^2+x-12}{x^2-ax+b}$

Note that $x^2+x-12$ also has $3$ as a root, so for an asymptote to exist we require $3$ to be a repeated root. Hence $x^2-ax+b=(x-3)^2\implies a=6,\,b=9$.

You can try also like that: Since $b= 3a-9$ we have $$f(x) = {(x+4)(x-3)\over x^2-ax+3a-9}= {(x+4)(x-3)\over (x-3)(x+3-a)} ={x+4\over x+3-a}$$

so $3-a=-3\implies a = 6$ and $b= 9$.