Answer
$$y' = \frac{{{e^x} + 2{e^{2x}} + 3{e^{3x}} + 2{e^{4x}} + {e^{5x}}}}{{\left( {1 + {e^x} + {e^{2x}}} \right)\left( {1 - {e^{3x}}} \right)}}$$
Work Step by Step
$$\eqalign{
& y = \ln \left( {\frac{{1 + {e^x} + {e^{2x}}}}{{1 - {e^{3x}}}}} \right) \cr
& {\text{quotient rule for logarithms ln}}\left( {\frac{u}{v}} \right) = \ln u - \ln b \cr
& y = \ln \left( {1 + {e^x} + {e^{2x}}} \right) - \ln \left( {1 - {e^{3x}}} \right) \cr
& {\text{find the derivative}} \cr
& y' = \left( {\ln \left( {1 + {e^x} + {e^{2x}}} \right)} \right)' - \left( {\ln \left( {1 - {e^{3x}}} \right)} \right)' \cr
& {\text{use }}\left( {\ln u} \right)' = \frac{{u'}}{u}, \cr
& y' = \frac{{\left( {1 + {e^x} + {e^{2x}}} \right)'}}{{1 + {e^x} + {e^{2x}}}} - \frac{{\left( {1 - {e^{3x}}} \right)'}}{{1 - {e^{3x}}}} \cr
& {\text{compute derivatives}} \cr
& y' = \frac{{{e^x} + 2{e^{2x}}}}{{1 + {e^x} + {e^{2x}}}} - \frac{{ - 3{e^{3x}}}}{{1 - {e^{3x}}}} \cr
& {\text{simplifying}} \cr
& y' = \frac{{{e^x} + 2{e^{2x}}}}{{1 + {e^x} + {e^{2x}}}} + \frac{{3{e^{3x}}}}{{1 - {e^{3x}}}} \cr
& y' = \frac{{{e^x} + 2{e^{2x}} - {e^{4x}} - 2{e^{5x}} + 3{e^{3x}} + 3{e^{4x}} + 3{e^{5x}}}}{{\left( {1 + {e^x} + {e^{2x}}} \right)\left( {1 - {e^{3x}}} \right)}} \cr
& y' = \frac{{{e^x} + 2{e^{2x}} + 3{e^{3x}} + 2{e^{4x}} + {e^{5x}}}}{{\left( {1 + {e^x} + {e^{2x}}} \right)\left( {1 - {e^{3x}}} \right)}} \cr} $$
