Answer
The curve is concave up when $\frac{d^{2}y}{dx^{2}} \gt 0 $, when $t \gt \frac{3}{2}$
Work Step by Step
$x=e^{t}$, $y=te^{-t}$
Step 1
$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}= \frac{-te^{-t}+e^{-t}}{e^{t}}=e^{-2t}(1-t)$
Step 2
Find second derivative
$\frac{d^{2}y}{dx^{2}}= \frac{\frac{d}{dt}(\frac{dy}{dx})}{\frac{dx}{dt}}= \frac{e^{-2t}(-1)+(1-t)(-2e^{-2t})}{e^{t}}=\frac{e^{-2t}(-1-2+2t)}{e^{t}}= e^{-2t}(2t-3)$
The curve is concave up for $\frac{d^{2}y}{dx^{2}} \gt 0$. This happens when $(2t-3)\gt 0$ or when $t \gt \frac{3}{2}$
