Answer
$ 0
Work Step by Step
Given
$$x=t-\ln t, \ \ \ \ y=t+\ln t $$
Since
\begin{aligned}
\frac{dy}{dt}& = 1+\frac{1}{t}\\
\frac{dx}{dt}& = 1-\frac{1}{t}\\
\end{aligned}
Then
\begin{aligned}
\frac{d y}{d x}&=\frac{d y / d t}{d x / d t}\\
&=\frac{1+1 / t}{1-1 / t}\\
&=\frac{t+1}{t-1}
\end{aligned}
and
\begin{aligned}
\frac{d^2 y}{d x^2}&=\frac{\frac{d}{d t}\left(\frac{d y}{d x}\right)}{d x / d t}\\
&=\frac{\frac{(t-1)(1)-(t+1)(1)}{(t-1)^2}}{(t-1) / t}\\
&=\frac{-2 t}{(t-1)^3}
\end{aligned}
Since the curve is concave up when $ \frac{d^2y}{dx^2}>0 $ ,then
$ \frac{d^2y}{dx^2}>0 $ for $ 0