Answer
$\displaystyle \frac{\partial u}{\partial x}=y \sin^{-1}(yz)$,
$\displaystyle \frac{\partial u}{\partial y}=\frac{xyz}{\sqrt{1-y^{2}z^{2}}}+x\sin^{-1}(yz)$,
$ \displaystyle \frac{\partial u}{\partial z}=\frac{xy^{2}}{\sqrt{1-y^{2}z^{2}}}$
Work Step by Step
Treat y and z as constant to calculate $\displaystyle \frac{\partial u}{\partial x}$
$\displaystyle \frac{\partial u}{\partial x} =y \sin^{-1}(yz)\cdot(1)=y \sin^{-1}(yz)$
Treat x and z as constant to calculate $\displaystyle \frac{\partial u}{\partial y}$
$\displaystyle \frac{\partial u}{\partial y}\stackrel{\text{product rule} }{=} x\{y \displaystyle \cdot\frac{\partial}{\partial y}[\sin^{-1}(yz)]+\frac{\partial}{\partial y}(y)\cdot \sin^{-1}(yz)\}$
... chain rule for the first term,
$=xy\displaystyle \cdot\frac{1}{\sqrt{1-(yz)^{2}}}(z) +x \cdot\sin^{-1}(yz)$
$=\displaystyle \frac{xyz}{\sqrt{1-y^{2}z^{2}}}+x\sin^{-1}(yz)$
Treat x and y as constant to calculate $\displaystyle \frac{\partial u}{\partial z}$
$ \displaystyle \frac{\partial u}{\partial z}=xy\cdot\frac{\partial}{\partial z}[\sin^{-1}(yz)]\qquad$... chain rule ...
$=xy\displaystyle \cdot\frac{1}{\sqrt{1-(yz)^{2}}}\cdot[\frac{\partial}{\partial z}(yz)]$
$=xy\displaystyle \cdot\frac{1}{\sqrt{1-(yz)^{2}}}\cdot[y]$
$=\displaystyle \frac{xy^{2}}{\sqrt{1-y^{2}z^{2}}}$
