Question 366019
The answer given by the other tutor is wrong.  He claims that the 2nd derivative of y with respect to x is the same as the 2nd derivative of y wrt t OVER the 2nd derivative of x wrt t.  THAT IS NOT TRUE!
Now {{{d^2y/dx^2 = d(dy/dx)/dx =( d(dy/dx)/dt)/(dx/dt)}}}.  Now {{{dy/dt = 1-1/t = (t-1)/t}}}, and {{{dx/dt = 1+1/t = (t+1)/t}}}.  Hence {{{dy/dx = ((t-1)/t)/((t+1)/t) = (t-1)/(t+1)}}}.  Hence
{{{d^2y/dx^2 = (d((t-1)/(t+1))/dt)/((t+1)/t) = (2/(t+1)^2)/((t+1)/t) = (2t)/(t+1)^3}}}. (Used the quotient rule on the numerator derivative!)
 For the parametric curve to be concave upward, {{{d^2y/dx^2 >0}}}, or {{{(2t)/(t+1)^3 > 0}}}.  Solving this inequality, the critical numbers of the inequality are 0 and -1.  Using the test numbers -2, -1/2, and 1, and checking for signs, the solution set is ({{{-infinity}}}, -1)U(0, {{{infinity}}}).