The equation is $\displaystyle y = x^2 e^{-x}$ with the point $\displaystyle (1, 1/e)$

I used the chain rule (I hope I used it correctly)

$\displaystyle f(x) = x^2$ , $\displaystyle f'(x) = 2x$

$\displaystyle g(x) = e^{-x}$ , $\displaystyle g'(x) = e^{-x}$

$\displaystyle y' = 2(e^{-x}) e^{-x}$

The problem is I don't know how to find the equation for the tangent line to the curve at the given point.