Answer
$${f_{{\text{ave}}}} = \frac{\pi }{{6\ln 3}}$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \frac{{{e^{3x}}}}{{1 + {e^{6x}}}};\,\,\,\left[ { - \frac{{\ln 3}}{6},0} \right] \cr
& {\text{Find the average value using the definition }}{f_{{\text{ave}}}} = \frac{1}{{b - a}}\int_a^b {f\left( x \right)dx} \cr
& {f_{{\text{ave}}}} = \frac{1}{{0 - \left( { - \frac{{\ln 3}}{6}} \right)}}\int_{ - \frac{{\ln 3}}{6}}^0 {\frac{{{e^{3x}}}}{{1 + {e^{6x}}}}dx} \cr
& {f_{{\text{ave}}}} = \frac{6}{{3\ln 3}}\int_{ - \left( {\ln 3} \right)/6}^0 {\frac{{3{e^{3x}}}}{{1 + {{\left( {{e^{3x}}} \right)}^2}}}dx} \cr
& {\text{Integrating }} \cr
& {f_{{\text{ave}}}} = \frac{6}{{3\ln 3}}\left( {{{\tan }^{ - 1}}{e^{3x}}} \right)_{ - \left( {\ln 3} \right)/6}^0 \cr
& {f_{{\text{ave}}}} = \frac{6}{{3\ln 3}}\left( {{{\tan }^{ - 1}}{e^{3\left( 0 \right)}} - {{\tan }^{ - 1}}{e^{ - 3\left( {\left( {\ln 3} \right)/6} \right)}}} \right) \cr
& {\text{Simplifying}} \cr
& {f_{{\text{ave}}}} = \frac{6}{{3\ln 3}}\left( {{{\tan }^{ - 1}}1 - {{\tan }^{ - 1}}{e^{ - \left( {\ln 3} \right)/2}}} \right) \cr
& {f_{{\text{ave}}}} = \frac{6}{{3\ln 3}}\left( {{{\tan }^{ - 1}}1 - {{\tan }^{ - 1}}{e^{ - \left( {\ln \sqrt 3 } \right)}}} \right) \cr
& {f_{{\text{ave}}}} = \frac{6}{{3\ln 3}}\left( {{{\tan }^{ - 1}}1 - {{\tan }^{ - 1}}\left( {\frac{1}{{\sqrt 3 }}} \right)} \right) \cr
& {f_{{\text{ave}}}} = \frac{6}{{3\ln 3}}\left( {\frac{\pi }{4} - \frac{\pi }{6}} \right) \cr
& {f_{{\text{ave}}}} = \frac{6}{{3\ln 3}}\left( {\frac{\pi }{{12}}} \right) \cr
& {f_{{\text{ave}}}} = \frac{\pi }{{6\ln 3}} \cr} $$
