Answer
$$y' = - {3^{ - x}}\ln 3$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = {3^{ - x}} \cr
& y = {3^{ - x}} \cr
& {\text{take logarithm natural on both sides}} \cr
& \ln y = \ln {3^{ - x}} \cr
& {\text{logarithm properties}} \cr
& \ln y = - x\ln 3 \cr
& {\text{differentiate}} \cr
& \left( {\ln y} \right)' = \left( { - x\ln 3} \right)' \cr
& \left( {\ln y} \right)' = - \ln 3\left( x \right)' \cr
& \frac{{y'}}{y} = - \ln 3 \cr
& y' = y\left( { - \ln 3} \right) \cr
& {\text{replace }}y = {3^{ - x}} \cr
& y' = {3^{ - x}}\left( { - \ln 3} \right) \cr
& {\text{simplify}} \cr
& y' = - {3^{ - x}}\ln 3 \cr} $$