First find dy/dt by taking the derivative of y=t+lnt
This gives you dy/dt=1+(1/t)
Then find dx/dt by taking the derivative of x=t-lnt
This gives you dx/dt=1-1/t
Now to find the two do (dy/dt)/(dx/dt) which gives you (1+1/t)/(1-1/t) which can be simplified to t+1/t-1 by multiplying the fraction by t/t
Now to find the second derivative, do d/dx[dy/dx]/dx/dt
d[dy/dx] can be found by applying the quotient rule to the dy/dx we found in part a which will give us
-2/(t-1)^2
Now divide this by dx/dt to get -2t/((t-1)(t-1)^2). Now we need to find our critical numbers which occur at where the second derivative is zero or undefined. Which in this case is at t=0 and t=1. If you plug values before in between t = 0, you will notice that the second derivative is positive here, and after t=1 it is negative.
Therefore, the curve is concave up on the interval (0,1)