Answer
$f_{xzy}=f_{yxz}=6yz^{2}$
Work Step by Step
Assuming that the third partial derivatives of $f$ are continuous, we apply Clairaut's theorem, ($f_{xzy}=f_{yxz}$).
The easiest first partial derivative is $f_{y}$, (the arcsin term is treated as constant)
$f_{y}=2xyz^{3}+0=2xyz^{3}$
next, partially derive by x,
$f_{yx}=2(1)yz^{3}=2yz^{3}$,
and lastly, by z
$f_{yxz}=2y(3z^{2})=6yz^{2}$
$f_{xzy}=f_{yxz}=6yz^{2}$
