Answer
$$6$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {6,2} \right)} \frac{{{x^2} - 3xy}}{{x - 3y}} \cr
& {\text{evaluate using the theorem 12}}{\text{.1}}{\text{, substitute }}6{\text{ for }}x{\text{ and }}2{\text{ for }}y{\text{ into }}f\left( {x,y} \right) \cr
& {\text{then}} \cr
& \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {6,2} \right)} \frac{{{x^2} - 3x{y^2}}}{{x - 3y}} = \frac{{{{\left( 6 \right)}^2} - 3\left( 6 \right)\left( 2 \right)}}{{6 - 3\left( 2 \right)}} \cr
& {\text{simplifying}} \cr
& \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {6,2} \right)} \frac{{{x^2} - 3x{y^2}}}{{x - 3y}} = \frac{0}{0}{\text{ indeterminate}} \cr
& {\text{simplify the expression }}\frac{{{x^2} - 3xy}}{{x - 3y}}{\text{ factoring the numerator}} \cr
& = \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {6,2} \right)} \frac{{x\left( {x - 3y} \right)}}{{x - 3y}} \cr
& = \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {6,2} \right)} x \cr
& {\text{evaluate using the theorem 12}}{\text{.1}} \cr
& = 6 \cr} $$
