Let's rewrite the equation as
e^(-x)+x-1/3=0
We can show that the function f(x)=e^(-x)+x-1/3 has a global minimum whose value is positive and so the former equation f(x)=0 has no real solutions.
To find the extrema of the function we look at the boundary of the domain (in this case, the reals RR) and to the stationary points.
f is continuous on RR and lim_{x to pm infty} f(x)=+infty. This means that f has no global maximum and at least one global minimum.
f'(x)=-e^[-x]+1 is null when e^(-x)=1, i.e. there's only one stationary point in x=0. This has to be the global minimum, because the extrema are always stationary points and we already guaranteed the existence of a global minimum.
min_{x in RR} f(x) = f(0)=e^0+0-1/3=1-1/3=2/3>0, so the function f is always positive and never crosses the x-axis.
We can conclude that f(x)=0 doesn't have real solutions.