Answer
\[\frac{dy}{dx}=(1-t)e^{-2t}\]
\[\frac{d^2y}{dx^2}=e^{-3t}(2t-3)\]
\[t>\frac{3}{2}\]
Work Step by Step
\[x=e^t\;,\;y=te^{-t}\]
Differentiate $x$ with respect to $t$:
\[\frac{dx}{dt}=e^{t}\;\;\;\;\;\ldots (1)\]
Differentiate $y$ with respect to $t$:
\[\frac{dy}{dt}=(1)e^{-t}-te^{-t}=(1-t)e^{-t}\]
\[\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{(1-t)e^{-t}}{e^{t}}\]
\[\Rightarrow \frac{dy}{dx}=(1-t)e^{-2t}\]
Differeniate $\displaystyle\frac{dy}{dx}$ with respect to $x$ using chain rule:
\[\frac{d^2y}{dx^2}=\frac{d}{dt}\left(\frac{dy}{dx}\right)\cdot\frac{dt}{dx}\]
\[\frac{d^2y}{dx^2}=\frac{d}{dt}\left[(1-t)e^{-2t}\right]\cdot\frac{dt}{dx}\]
\[\frac{d^2y}{dx^2}=\left[-e^{-2t}-2(1-t)e^{-2t}\right]\cdot\frac{dt}{dx}\]
\[\Rightarrow \frac{d^2y}{dx^2}=\left[-e^{-2t}(3-2t)\right]\cdot\frac{dt}{dx}\]
Using (1)
\[\Rightarrow \frac{d^2y}{dx^2}=\left[-e^{-2t}(3-2t)\right]\cdot(e^{-t})\]
\[\Rightarrow \frac{d^2y}{dx^2}=-e^{-3t}(3-2t)\]
\[\Rightarrow \frac{d^2y}{dx^2}=e^{-3t}(2t-3)\]
$\displaystyle \frac{d^2y}{dx^2}>0$ if $t>\displaystyle\frac{3}{2}$ so The curve is concave up if $t>\displaystyle\frac{3}{2}$.
