Well, since it requires a line through the y-intercept, it would be a good idea to find that! The y-intercept of a graph is where it crosses the y-axis: where x= 0. Now assuming you meant

which is NOT what you wrote (what you wrote was

) but does have the property that its derivative is still

, then since [itex]f(0)= \frac{1}{2}e[/itex], the y-intercept is

. The slope of the tangent line to the graph is the derivative evaluated at x= 0 which is again [tex]\frac{1}{2}e[/itex]. Do you know how to find the slope and equation of the perpendicular line from that?