Determine the point at which the tangent to the function f(x)=(e^2)/(x^3) is horizontal

Answer 1

f'(x) = -3e^2/x^4
f' is never zero, so neither is the tangent.
Now, if you meant
f(x) = (e^x^2)/x^3
then f'(x) = (e^x^2)/x^4 (2x^2 - 3)
and that is zero when
2x^2 - 3 = 0
which I'm sure you can solve.