Answer
The equation of tangent at (0, 1) is $y=e$.
Work Step by Step
The slope of the tangent can be calculated by taking first derivative of the function
$y=\frac{e^{x}}{x}$
$y'=\frac{xe^{x}-e^{x}}{x^{2}}=\frac{e^{x}(x-1)}{x^{2}}$
Slope of the tangent at $(1,e)$ is given as follows:
$y'|_{(1,e)}=0$
The equation of tangent at (0, 1) is
$(y-y_{1})=m(x-x_{1})$
Here $ m=0$
$(y-e)=0(x-1)$
Hence, $y=e$