Don't feel bad that you think things are confusing; In my opinion, this is a rather subtle problem for a second year calculus course.
It is the right hand side of the inequality being considered here. The maximum value of $xyz$ is calculated in the special case that the sum $x+y+z$ is known: the formulation given in the hint is finding the maximum value of the function $\tag{1}g(x,y,z)=xyz$ subject to to constraints $x\ge0,y\ge0,z\ge0$ and $x+y+z=A$, where $A$ is a fixed, but otherwise arbitrary, number. The last constraint forces $z=A-x-y$, which allows us to phrase the maximization problem as
Find the maximum value of $\tag{2} f(x,y)=xy(A-x-y) $ over the region defined by $x,y\ge0$, $A-x-y\ge 0$. Note that that last inequality can be written as $A\ge x+y$. This region can be described as given in the hint "the region enclosed by $x=0$, $y=0$, and $x+y=A$".
Why do this? Well, .... umm ... it just works:
So, you go about finding the gradient of $f$, setting it equal to the zero vector, finding the maximum value of $f$ over those zeroes and at the boundary of the region, and then find the largest of those maximums and where it occurs. (I can provide details here if you like, but the calculations aren't too bad.)
You will find that the maximum value of $f$ over the given region is attained at $x=y={A\over3}$. Thus, the maximum value of $g$ is attained at $x=y=z={A\over 3}$ (remember $A=x+y+z)$ and this maximum value is $\textstyle g\bigl({A\over3},{A\over3},{A\over3}\bigr)={A\over 3}\cdot{A\over3}\cdot{A\over3}={A^3\over 27}.$
So, what have we shown?
We have shown the following:
If $x,y,z\ge0$ and if $x+y+z=A$ (where $A$ is now just the sum of the given $x,y,z$), then $xyz\le {A^3\over 27}$. Thus, we have $ xyz\le{A^3\over27}={(x+y+z)^3\over 27}; $ and taking cube roots of both sides of the above gives your inequality (which, by the way, is a very important and famous inequality known as the Arithmetic Mean-Geometric Mean, or AM-GM, inequality).