When I tried this problem I simply used $1$ as the upper bound for the integral over $x$. Afterwards, I read the solution shown in the image and realized that my answer was wrong. My reasoning was that by using $1$ as the upper bound, I was essentially describing a different shape - a rectangular prism. However, this reasoning seems flimsy and unfinished. Could someone give a better reason as to why the upper bound needed to be $1-y$?
Note that the integral over $x$ does not have constant bounds when we integrate with respect to $dx\,dy\,dz$. You should be able to gather that by staring at the picture, selecting different values of $y$, and running your eye from the $y$-axis to the boundary of the region. If we plot that region in the $x,y$-plane without the extra $z$-axis, you should see that the boundary of the region is a segment of the line defined by $y = 1 - x$. That is, the $y$-intercept of that line is $y = 1$ and the slope of the line segment is $-1$. Hence the bounds on the integral over $x$ are $x=0$ to $x=1-y$.
You would use the bounds $x = 0$ to $x = 1$ if you were integrating with respect to $dy\,dx\,dz$, for instance. Then your bounds on $y$ would be $y = 0$ to $y = 1 - x$ by the same reasoning I gave above.