Answer
$$2$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {2,0} \right)} \frac{{{x^2} - 3x{y^2}}}{{x + y}} \cr
& {\text{use the law 5 }}\mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {a,b} \right)} \frac{{f\left( {x,y} \right)}}{{g\left( {x,y} \right)}}.\,\,\,{\text{theorem 12}}{\text{.2 }}\left( {{\text{see page 887}}} \right){\text{. then}} \cr
& = \frac{{\mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {2,0} \right)} \left( {{x^2} - 3x{y^2}} \right)}}{{\mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {2,0} \right)} \left( {x + y} \right)}} \cr
& {\text{use law 1}} \cr
& = \frac{{\mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {2,0} \right)} \left( {{x^2}} \right) - \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {2,0} \right)} \left( {3x{y^2}} \right)}}{{\mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {2,0} \right)} \left( x \right) + \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {2,0} \right)} \left( y \right)}} \cr
& {\text{law constant multiply}} \cr
& = \frac{{\mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {2,0} \right)} \left( {{x^2}} \right) - 3\mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {2,0} \right)} \left( {x{y^2}} \right)}}{{\mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {2,0} \right)} \left( x \right) + \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {2,0} \right)} \left( y \right)}} \cr
& {\text{evaluate using the theorem 12}}{\text{.1}} \cr
& = \frac{{{{\left( 2 \right)}^2} - 3\left( 2 \right){{\left( 0 \right)}^2}}}{{2 + \left( 0 \right)}} \cr
& {\text{simplifying}} \cr
& = 2 \cr} $$
