Since the derivative of a function #f#, denoted #f'#, is the "instantaneous rate of change" of the function. A way to tell the behavior of a function around a point is to take the derivative. If #f'>0# then #f# is increasing. If #f'<0# then #f# is decreasing.
Since #x^(-r)=1/x^r#, we can say
#f(x)=-x/(e^(x^2-3x+2))=-x(e^(-x^2+3x-2))#
And since the Product rule states
#[h(x)g(x)]'=h'(x)g(x)+h(x)g'(x)#
we can say
#f'(x)=-e^(-x^2+3x-2)-x(e^(-x^2+3x-2))'#
and since the chain rule states #(h(g(x)))'=h'(g(x))g'(x)#
we can say
#(e^(-x^2+3x-2))'=(-x^2+3x-2)'e^(-x^2+3x-2)#
#=-2x+3(e^(-x^2+3x-2))=-2xe^(-x^2+3x-2)+3e^(-x^2+3x-2)#
Then
#f'(x)=-e^(-x^2+3x-2)+2x^2e^(-x^2+3x-2)-3xe^(-x^2+3x-2)#
Now to check for the point #x=4# plug in #4#
Then
#f'(4)=-e^(-16+12-2)+32e^(-16+12-2)-12e^(-16+12-2)#
#=-e^-6+32e^-6-12e^-6=19e^-6#
and since #f'(4)=19e^-6# then #f'(4)>0# #therefore# #f# is increasing at the point #x=4#
