How would I go about drawing the graph of f(x) = sin(x)/(e^x) for -π≤x≤2π?

Drawing this graph uses the same techniques as with most other functions. Just from inspection, we can see that as x increases, e^x also increases, and so the amplitude of the function will decrease (and tend towards zero) due to the inversely proportional relationship. From general knowledge of the sine function, we can also see that sin(x) = 0 at -π, 0, π and 2π, and so f(x) will too equal zero at these points.
Differentiating the function using the quotient rule gives us f'(x) = (cos(x) - sin(x))/(e^x), and we can find the minima and maxima by setting this equal to zero and getting sin(x) = cos(x). Dividing through by cos(x) gives tan(x) = 1, which we know occurs at the points x = π/4 + nπ. In the interval given, these minima/maxima are at -3π/4, π/4 and 5π/4, and so we can calculate f(x) at these points to be -7.46, 0.32 and -0.01 respectively. Plot these minima and maxima along with the points where the graph intercepts the axes, and draw the graph connecting these points (will look like a damped sine wave).