Answer
$6yz^2$
Work Step by Step
Given: $xy^2z^3+ \sin^{-1} (x \sqrt x)$
Differentiating $xy^2z^3+ \sin^{-1} (x \sqrt x)$ partially times with respect to $y$ keeping $x$ and $z$ constant .
$f_y=\dfrac{∂[xy^2z^3+ \sin^{-1} (x \sqrt x)]}{∂y}=x2yz^3$
Differentiating the above equation partially with respect to $x$ keeping $y$ and $z$ constant .
$f_yx=\dfrac{∂[x2yz^3]}{∂y}=2yz^3$
and
$f_{xyz}=6yz^2$
Hence, $f_{xyz}=f_{xzy}=6yz^2$
