Answer
$$y=2x-e$$
Work Step by Step
Given $$y=x\ln x,\ \ \ (e,e)$$
Since
$$y'=x\frac{1}{x}+\ln x=1+\ln x$$
Then $m=y'\bigg|_{(e,e)}=2$, hence the tangent line given by
\begin{align*}
\frac{y-y_1}{x-x_1}&=m\\
\frac{y-e}{x-e}&=2\\
y-e&=2x-2e\\
y&=2x-e
\end{align*}
