This equation is a very short question that has a very long answer,
Let me show you how to get the answer,
g(x) = x/e^x
One thing you can do is find the first few derivatives, and then you can validate the solution by induction.
g'(x) = [ (1)e^x - x(e^x) ] / (e^x)^2
= [e^x - xe^x] / [e^x]^2
= (e^x)(1 - x) / (e^x)^2
= (1 - x)/e^x
g''(x) = [ (-1)(e^x) - (1 - x)(e^x) ] / (e^x)^2
= [ -e^x - e^x + x e^x ] / (e^x)^2
= [ -2e^x + x e^x ] / (e^x)^2
= (e^x)(-2 + x) / (e^x)^2
= (-2 + x) / (e^x)
= (-1)(2 - x)/e^x
g'''(x) = [ (1)(e^x) - (-2 + x)(e^x) ] / (e^x)^2
g'''(x) = [ e^x - (-2e^x + x e^x) ] / (e^x)^2
g'''(x) = [ e^x + 2e^x - x e^x ] / (e^x)^2
g'''(x) = [ 3e^x - x e^x ] / (e^x)^2
g'''(x) = (e^x)(3 - x) / (e^x)^2
g'''(x) = (3 - x)/e^x
By inspection, it appears the pattern is
g^(n)(x) = (-1)^(n - 1) (n - x)/e^x