Answer
$$y' = \frac{{x{e^{ - x}} - {e^{ - x}}}}{{1 - x{e^{ - x}}}}$$
Work Step by Step
$$\eqalign{
& y = \ln \left( {1 - x{e^{ - x}}} \right) \cr
& {\text{differentiate}} \cr
& y' = \left( {\ln \left( {1 - x{e^{ - x}}} \right)} \right)' \cr
& {\text{use }}\left( {\ln u} \right)' = \frac{{u'}}{u} \cr
& y' = \frac{{\left( {1 - x{e^{ - x}}} \right)'}}{{1 - x{e^{ - x}}}} \cr
& {\text{product rule}} \cr
& y' = \frac{{ - \left( x \right)\left( {{e^{ - x}}} \right)' - \left( {{e^{ - x}}} \right)\left( x \right)'}}{{1 - x{e^{ - x}}}} \cr
& y' = \frac{{ - \left( x \right)\left( { - {e^{ - x}}} \right) - \left( {{e^{ - x}}} \right)\left( 1 \right)}}{{1 - x{e^{ - x}}}} \cr
& {\text{simplify}} \cr
& y' = \frac{{x{e^{ - x}} - {e^{ - x}}}}{{1 - x{e^{ - x}}}} \cr} $$
