Construct a rational polynomial function such that it has zeroes at $x= -2 , 3$ and has vertical asymptotes at $x= 2, -3$ and has a oblique asymptote $y= x-5$.
I found this one
$$=\frac{(x+2)(x-3)}{(x-2)(x+3)}\\ =\frac{x^2-x-6}{x^2+x-6}$$
then I have no idea what to do.
To ensure all of the required properties, consider $$f(x)=x-5+\frac{p(x)}{(x-2)(x+3)}\,,$$ which has asymptotes in the right places.
To get the zeros at $3$ and $-2$, we need $f(3)=0$ and $f(-2)=0$. These conditions imply $p(3)=12$ and $p(-2)=-28$. Let $p(x)=ax+b$. Then we have $3a+b=12$ and $-2a+b=-28$. Solving simultaneously, we obtain $a=8,b=-12$, so we have $$f(x)=x-5+\frac{8x-12}{(x-2)(x+3)}\,.$$