$\displaystyle x = e - e^t$

$\displaystyle y = t e^{-t}$

Find $\displaystyle \frac{dy}{dx}$ and $\displaystyle \frac{d^2y}{dx^2}$ and state where the graph is concave upward and downward.

I found dy/dx already: $\displaystyle \frac{1-e^{-t}}{1-e^t}$

But when I used the quotient rule to find $\displaystyle \frac{\frac{d}{dt}(\frac{dy}{dx})}{\frac{dx}{dt}}$ I got $\displaystyle \frac{e^{-t}-1+e^t-1}{(1-e^t)^3}$

I think this is right, but I might have entered it wrong. Is it?

I know the concavity is based on when the second derivative is positive and negative.