Evaluating the innermost integral gives us $\displaystyle (x+2y)z\big|_0^x = x^2+2xy$.
Now the integral is $\displaystyle \int_0^1\int_{x^2}^x(x^2+2xy)\,dy\,dx$
Evaluating the next part of the integral gives us $\displaystyle (x^2y+xy^2)\big|_{x^2}^x = x^3+x^3-x^4-x^5 = 2x^3-x^4-x^5$
Now the integral is $\displaystyle \int_0^1 (2x^3-x^4-x^5)\,dx$
Evaluating this gives us $\displaystyle \left(\frac{x^4}{2}-\frac{x^5}{5}-\frac{x^6}{6}\right)\bigg|_0^1 = \frac{1}{2}-\frac{1}{5}-\frac{1}{6} = \frac{2}{15}$
So I think $\displaystyle \frac{2}{15}$ is the answer.