\(\displaystyle y = t e^{-t}\)

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.