Answer
$$\eqalign{
& a.\,\mathop {\lim }\limits_{x \to \infty } f\left( x \right) = 1;\,\,\,\mathop {\lim }\limits_{x \to - \infty } f\left( x \right) = 1;\,\,\,\,\,\,y = 1 \cr
& b.\,x = 0;\,\,\,\,\,\mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) = - \infty ;\,\,\,\,\,\mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = + \infty \cr} $$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \frac{{{x^2} - 9}}{{x\left( {x - 3} \right)}} \cr
& {\text{Factor the numerator}} \cr
& f\left( x \right) = \frac{{\left( {x + 3} \right)\left( {x - 3} \right)}}{{x\left( {x - 3} \right)}} \cr
& f\left( x \right) = \frac{{x + 3}}{x} \cr
& f\left( x \right) = 1 + \frac{3}{x} \cr
& \cr
& a.{\text{ Calculate }}\mathop {\lim }\limits_{x \to \infty } f\left( x \right){\text{ and }}\mathop {\lim }\limits_{x \to - \infty } f\left( x \right) \cr
& *\,\,\mathop {\lim }\limits_{x \to \infty } f\left( x \right) \cr
& \mathop {\lim }\limits_{x \to \infty } \left( {1 + \frac{3}{x}} \right) \cr
& {\text{Use limits properties}} \cr
& = \mathop {\lim }\limits_{x \to \infty } \left( 1 \right) + \overbrace {\mathop {\lim }\limits_{x \to \infty } \left( {\frac{3}{x}} \right)}^{{\text{approaches to 0}}} \cr
& = 1 + 0 \cr
& = 1 \cr
& and\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \cr
& *\,\,\mathop {\lim }\limits_{x \to - \infty } f\left( x \right) \cr
& \mathop {\lim }\limits_{x \to - \infty } \left( {1 + \frac{3}{x}} \right) \cr
& = \mathop {\lim }\limits_{x \to - \infty } \left( 1 \right) + \overbrace {\mathop {\lim }\limits_{x \to - \infty } \left( {\frac{3}{x}} \right)}^{{\text{approaches to 0}}} \cr
& \,\, = 1 + 0 \cr
& \,\, = 1 \cr
& {\text{Then the horizontal asymptote of }}f\left( x \right){\text{ is }}y = 1 \cr
& \cr
& b.\,\,{\text{The function is not defined for }}x = 0,{\text{ then}} \cr
& \,\,\,\,\,{\text{there is a vertical asymptotes at }}x = 0 \cr
& \cr
& {\text{Analyze }}\mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right){\text{ and }}\mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) \cr
& \mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ - }} \left( {1 + \frac{3}{x}} \right) \cr
& {\text{ }} = \mathop {\lim }\limits_{x \to {0^ - }} \left( 1 \right) + \overbrace {\mathop {\lim }\limits_{x \to {0^ - }} \left( {\frac{3}{x}} \right)}^{{\text{approaches to - }}\infty } \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1 + \frac{3}{{{0^ - }}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1 - \infty \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = - \infty \cr
& \mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ + }} \left( {1 + \frac{3}{x}} \right) \cr
& {\text{ }} = \mathop {\lim }\limits_{x \to {0^ + }} \left( 1 \right) + \overbrace {\mathop {\lim }\limits_{x \to {0^ + }} \left( {\frac{3}{x}} \right)}^{{\text{approaches to + }}\infty } \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1 + \infty \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \infty \cr} $$