Solution
= Solution
Integrating the <mixed partial derivative> successively gives $z=x^3y^2/6+A(x)+B(y)$. The first boundary value gives $A(x)=x^2-1-B(0)$. Substitution at $x=1$ then gives $B(y)-B(0)=\sin y-y^2/6$. Therefore
$$
\boxed{z(x,y)=x^2-1+\sin y+\frac{(x^3-1)y^2}{6}.}
$$
Its mixed derivative is $x^2y$, and direct evaluation checks both boundary values. Their common corner is compatible: both prescribe $z(1,0)=0$.