To show you how far I'll go to avoid the quotient rule... :^)

First I notice that x^2 + x + 2 > 0 always (negative discriminant and positive at x=0). That means that y>0 for all x>0 and maximizing y is the same as minimizing 1/y.

1/y = (x^2 + x + 2)/x = x + 1 + 2/x

(1/y)' = 1 - 2/x^2

Solving for (1/y)' = 0 gets x^2 = 2 and only the positive square root of 2 fits the x>0 restriction.

That's problem 1, the easy way.

For problem 2, find the minimum value for fixed a,b and variable x.

y'(x) = 3x^2 - 3ab = 0

x^2 = ab

x = sqrt(ab)

The minimum y value on [0, oo) is:

y(sqrt(ab)) = (ab)^(3/2) - 3ab*sqrt(ab) + a^3 + b^3

= (ab)^(3/2) - 3*(ab)^(3/2) + a^3 + b^3

= a^3 - 2*a^(3/2)*b^(3/2) + b^3

That's the square of [a^(3/2) - b^(3/2)], and a square of a real number is never is never negative. The minimum value of y(x) on [0,oo) can't be negative, y(x) >= 0 for all x>=0.