Answer
$y=x-1$
$y=\frac{1}{e}$
Work Step by Step
$y=\frac{(\ln x)}{x}\\
y^{\prime}=\frac{(x)\left(\frac{1}{x}\right)-(\ln x)(1)}{x^{2}}\\
y^{\prime}=\frac{1-\ln x}{x^{2}}$
Gradient at point $(1,0)$
$y^{\prime}=\frac{1-\ln 1}{(1)^{2}}\\
y^{\prime}=1$
Tangent line at point $(1,0)$
$y-0=1(x-1)\\
y=x-1$
Gradient at point $(e,1/e)$
$y^{\prime}=1-\ln e\\
y^{\prime}=0$
Tangent line at point $(e,1/e)$
$y^{\prime}=\frac{1}{e}\\
y=\frac{1}{e}$