Answer
$$\frac{1}{4}$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{\left( {u,v} \right) \to \left( {8,8} \right)} \frac{{{u^{1/3}} - {v^{1/3}}}}{{{u^{2/3}} - {v^{2/3}}}} \cr
& {\text{Evaluating the limit}} \cr
& \mathop {\lim }\limits_{\left( {u,v} \right) \to \left( {8,8} \right)} \frac{{{u^{1/3}} - {v^{1/3}}}}{{{u^{2/3}} - {v^{2/3}}}} = \frac{{{{\left( 8 \right)}^{1/3}} - {{\left( 8 \right)}^{1/3}}}}{{{{\left( 8 \right)}^{2/3}} - {{\left( 8 \right)}^{2/3}}}} \cr
& \mathop {\lim }\limits_{\left( {u,v} \right) \to \left( {8,8} \right)} \frac{{{u^{1/3}} - {v^{1/3}}}}{{{u^{2/3}} - {v^{2/3}}}} = \frac{{{{\left( 8 \right)}^{1/3}} - {{\left( 8 \right)}^{1/3}}}}{{{{\left( 8 \right)}^{2/3}} - {{\left( 8 \right)}^{2/3}}}} = \frac{0}{0}{\text{Ind}} \cr
& {\text{Then,}} \cr
& \mathop {\lim }\limits_{\left( {u,v} \right) \to \left( {8,8} \right)} \frac{{{u^{1/3}} - {v^{1/3}}}}{{{{\left( {{u^{1/3}}} \right)}^2} - {{\left( {{v^{1/3}}} \right)}^2}}} \cr
& {\text{Factoring the denominator}} \cr
& = \mathop {\lim }\limits_{\left( {u,v} \right) \to \left( {8,8} \right)} \frac{{{u^{1/3}} - {v^{1/3}}}}{{\left( {{u^{1/3}} + {v^{1/3}}} \right)\left( {{u^{1/3}} - {v^{1/3}}} \right)}} \cr
& = \mathop {\lim }\limits_{\left( {u,v} \right) \to \left( {8,8} \right)} \frac{1}{{{u^{1/3}} + {v^{1/3}}}} \cr
& {\text{Evaluating}} \cr
& \mathop {\lim }\limits_{\left( {u,v} \right) \to \left( {8,8} \right)} \frac{1}{{{u^{1/3}} + {v^{1/3}}}} = \frac{1}{{{{\left( 8 \right)}^{1/3}} + {{\left( 8 \right)}^{1/3}}}} = \frac{1}{{2 + 2}} \cr
& \mathop {\lim }\limits_{\left( {u,v} \right) \to \left( {8,8} \right)} \frac{1}{{{u^{1/3}} + {v^{1/3}}}} = \frac{1}{4} \cr} $$
