Answer
$$y = \frac{{\cos e}}{e}x - \cos e + 1$$
Work Step by Step
$$\eqalign{
& y = {\left( {\ln x} \right)^{\cos x}} \cr
& {\text{Use logarithmic differentiation}} \cr
& \ln y{\text{ = ln}}\left( {{{\left( {\ln x} \right)}^{\cos x}}} \right) \cr
& \ln y = \left( {\cos x} \right)\ln \left( {\ln x} \right) \cr
& \frac{d}{{dx}}\left[ {\ln y} \right] = \frac{d}{{dx}}\left[ {\left( {\cos x} \right)\ln \left( {\ln x} \right)} \right] \cr
& \frac{1}{y}\frac{{dy}}{{dx}} = \cos x\frac{d}{{dx}}\left[ {\ln \left( {\ln x} \right)} \right] + \ln \left( {\ln x} \right)\frac{d}{{dx}}\left[ {\cos x} \right] \cr
& \frac{1}{y}\frac{{dy}}{{dx}} = \cos x\left( {\frac{1}{{x\ln x}}} \right) - \ln \left( {\ln x} \right)\sin x \cr
& \frac{{dy}}{{dx}} = y\left[ {\cos x\left( {\frac{1}{{x\ln x}}} \right) - \ln \left( {\ln x} \right)\sin x} \right] \cr
& {\text{Calculate the slope at the point }}\left( {e,1} \right) \cr
& m = {\left. {\frac{{dy}}{{dx}}} \right|_{\left( {e,1} \right)}} \cr
& m = \left( 1 \right)\left[ {\cos e\left( {\frac{1}{{e\ln e}}} \right) - \ln \left( {\ln e} \right)\sin e} \right] \cr
& m = \left( 1 \right)\left[ {\frac{{\cos e}}{e} - \ln \left( 1 \right)\sin e} \right] \cr
& m = \frac{{\cos e}}{e} \cr
& {\text{The equation of the tangent line at the point }}\left( {e,1} \right){\text{ is}} \cr
& y - {y_1} = m\left( {x - {x_1}} \right) \cr
& y - 1 = \frac{{\cos e}}{e}\left( {x - e} \right) \cr
& y - 1 = \frac{{\cos e}}{e}x - \cos e \cr
& y = \frac{{\cos e}}{e}x - \cos e + 1 \cr} $$
