Answer
$$y = - \frac{x}{e}$$
Work Step by Step
$$\eqalign{
& {\text{Let }}f\left( x \right) = \ln \left( { - x} \right){\text{ and the point }}{x_0} = - e \cr
& {\text{then }}f\left( {{x_0}} \right) = f\left( { - e} \right) = \ln \left( { - \left( { - e} \right)} \right) = 1 \cr
& {\text{we have the point}}\left( { - e,1} \right) \cr
& {\text{Find the derivative of }}f\left( x \right) \cr
& f'\left( x \right) = \left( {\ln \left( { - x} \right)} \right)' \cr
& f'\left( x \right) = \frac{{ - 1}}{{ - x}} \cr
& f'\left( x \right) = \frac{1}{x} \cr
& {\text{evaluate }}f'\left( x \right){\text{ at the point }}{x_0}{\text{ to find the slope}} \cr
& f'\left( { - e} \right) = - \frac{1}{e} \cr
& f'\left( { - e} \right) = - \frac{1}{e} \cr
& {\text{then }}m = - \frac{1}{e} \cr
& \cr
& {\text{using the equation of the point - slope form }}y - {y_1} = m\left( {x - {x_1}} \right),{\text{ point }}\left( { - e,1} \right) \cr
& y - \left( 1 \right) = - \frac{1}{e}\left( {x - \left( { - e} \right)} \right) \cr
& {\text{Simplify}} \cr
& y - 1 = - \frac{1}{e}\left( {x + e} \right) \cr
& y - 1 = - \frac{x}{e} - 1 \cr
& y = - \frac{x}{e} \cr} $$
