I'm going to try to expand on @SangchulLee's answer a little bit.
You think through the original problem by realizing that our 3-dimensional region projects down to the region in the $xy$-plane between the $y$-axis and $x=y^3$ [or, equivalently, $y=x^{1/3}$]. In $\Bbb R^3$, $x=y^3$ is a vertical cylinder over this plane curve. For each point in that plane region, a vertical line enters the region at $z=0$ and exits at $z=y^2$. So this surface is a parabolic cylinder, with lines parallel to the $x$-axis. Here's a sketch with the assistance of Mathematica:
Indeed, the region lies over the rectangle $R=[0,8]\times [0,4]$ in the $xz$-plane, and the two cylinders intersect along the crease $y=z^{1/2}=x^{1/3}$, which projects to the curve $C$ given by $x^2=z^3$ in the $xz$-plane. If we take a point in $R$, we want to know when a line parallel to the $y$-axis enters and exits our region. If we're below $C$ in $R$, then $z^3<x^2$ and the line enters at $y=x^{1/3}$ and exits at $y=2$. (As a check, if $y\ge x^{1/3}$, then $y\ge z^{1/2}$ as well in this range, so we're "inside" both cylinders.) If we're above $C$ in $R$, then $z^3>x^2$ and the line enters at $y=z^{1/2}$ and exits at $y=2$.
Thus, we end up with the following iterated integral in the order $dy\,dz\,dx$:
$$\int_0^8\int_0^{x^{2/3}}\int_{x^{1/3}}^2 dy\,dz\,dx + \int_0^8\int_{x^{2/3}}^4\int_{z^{1/2}}^2 dy\,dz\,dx.$$
(By the way, you should be able to set up the limits in the orders $dx\,dz\,dy$ or $dx\,dy\,dz$ much more easily.)
The region of the integral is characterized by
$$
0 \leq y \leq 2
\\ 0 \leq x \leq y^3 \\ 0 \leq z \leq y^2
$$
If we have in mind that $x$ should be the outermost variable, the first thing is to deduce that $$0 \leq x \leq y^3 \leq 2^3 = 8 \\ 0 \leq z \leq y^2 \leq 4$$
The $0 \leq x \leq 2^3 = 8$ part gives the outer integral limits.
And we also have two lower-boundary conditions on $y$, namely
$$ y \geq x^{1/3} \\ y \geq z^{1/2}$$
And here is why the integral has to be split up -- we have no clue whether $x^{1/3}$ is more or less than $z^{1/2}$. If it is greater or equal, then $z\leq x^{2/3}$ and the limits will be
$$
\int_{x=0}^8\int_{z= 0}^{x^{2/3}} \int_{y=x^{1/3}}^2
$$
and if $x^{1/3} < z^{1/2}$ the upper limit on $z$ is given by $z \leq y^2 \leq 4$ and the limits will be
$$
\int_{x=0}^8\int_{z= x^{2/3}}^{4} \int_{y=z^{1/2}}^2
$$
Best Answer
The bounds of integration determine equations that bound a solid $S$ in three-dimensional space. After you integrate with respect to the first variable, you should orthogonally project $S$ along the axis specified by that first variable onto the plane spanned by the other two variables. That projection then determines a two-dimensional region which is the domain over which you integrate next.
In your original integral, you have $$\int_{1}^{0}\int_{y}^{1}\int_{0}^{y} f(x,y,z)\,dz\,dx\,dy = -\int_{0}^{1}\int_{y}^{1}\int_{0}^{y} f(x,y,z)\,dz\,dx\,dy.$$ Note that I have switched the outer bounds of integration and changed the sign of the integral to alleviate the minor annoyance that $1$ is not less than $0$. Now, the bounds of integration imply three inequalities that specify the domain of integration. $$ \begin{array}{l} 0<z<y \\ y<x<1 \\ 0<y<1. \end{array} $$ Now, the solid, together with the projections of interest, looks like so:
Now, I guess the reason that the $yz$ and $xy$ projections are relatively easy is that those projections correspond to cross-sections of the solid, i.e. $z=0$ for the $xy$ projection and $x=1$ for the $yz$ projection. This allows us to just set the variable that we integrate with respect to first to that constant. It's even easier here, since neither $x$ nor $z$ appear explicitly in the bounds. We can't do that here but, when we think of it as a projection as we have here, we can see that it's just the triangle $$\left\{(x,z): 0<x<1 \text{ and } 0<z<x \right\}.$$