Find the x-coordinate of the stationary points of the curve and determine the nature of these stationary points. The equation of a curve is $y=x^2e^{-x}$.

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*Find the x-coordinate of the stationary points of the curve and determine the nature of these stationary points.

*Show that the equation of the normal to the curve at the point where $x=1$ is $e^2x+ey = 1+e^2$.


This is the full question I am having difficulty solving, I simply don't know where to begin. I moved back to my home country because of covid and now I am doing self-studying I don't know how to solve this any help wold be great and much appreciated.

$y=x^2e^{-x} \implies y'(x)=2xe^{-x}-x^2e^{-x} \implies y'(1)=e^{-1}.$
So the slope of normal at $x=1$ is $m=-1/y'(1)=-e$. So the equation of line having slope $-e$ and passing through the point $(1,e^{-1})$ is
$$y-e^{-1}=-e(x-1) \implies ey+e^2 x=1+e^2$$
