Verified answer

1) Your dy/dx is correct, and indeed this parametric function never has dy/dx=0 (in other words, it has no local minimum or maximum)

2) The question asks: if a function approaches a flat horizontal line at infinity, what happens to its derivative at infinity?

Well, you know that the derivative of a flat line is 0: y(x)=C → dy/dx=0

Another way to see that is to realize that if a function approaches a flat asymptote, it changes less and less and less as x tends to infinity. At infinity, it will completely stop changing (and the change of a function is its derivative; namely dy/dx will become 0)

In your case, if x→+∞, t also →+∞, and

lim {t→+∞} dy/dx = lim {t→+∞} 1/[ t(1+t)² ] = 1/+∞ = 0

1) dy/dx is zero only when x is negative infinity. It will not be zero for any real value of x. What you did is correct.

2) If there is a horizontal asymptote for y = f(x) at x = 1, that means that lim (x-> inf) y = 1. You mixed up y and dy/dx. To find lim (x-> inf) dy/dx, find the limit of its derivative that you have found earlier as x approaches infinity. You will notice that this will be 0. The limits of y and dy/dx are not the same.

Hope this helps.