My solution:
Integrate w.r.t $x$ and get $zx$, after inserting the boundaries of $x$ I get $zy$.
Integrate w.r.t $z$ and get $y \frac{z^2}{2}$, after inserting the boundaries of $z$ I get $\frac{3y^3-2y^2-y}{2}$.
Integrate w.r.t $y$ and get $\frac{3y^4}{8} - \frac{y^3}{3} - \frac{y^2}{4}$, after inserting the boundaries of $y$ I get $\frac{-5}{24}$.
Is my solution correct?
Yes, your answer is correct. You have explained every step very clearly. Good Job!