Let r(t)=(t,e^-t); write the acceleration vector at time t=0 as a sum of tangential (i.e., parallel to the velocity vector) and normal (i.e., orthogonal) components.
Your post is a bit confusing, but I assume you mean:
r(t) is a parameterized curve such that x(t) = t, y(t) = e^-t.
Then r'(t) = v(t) is parameterized by x'(t) = 1, y'(t) = -e^-t,
and r''(t) = a(t) is parameterized by x''(t) = 0, y''(t) = e^-t.
Then at t = 0, v(0) = (1,-1), a(0) = (0,1). However, if you want a(0) in terms of the velocity vector (1,-1) and an orthogonal vector (1,1), you can do this by eyeball and say a(0) = -(1/2)*(-1,1) + (1/2)*(1,1).