Answer
$$\frac{\partial}{\partial x}u=\frac{y}{z}x^\frac{y-z}{z}.$$
$$\frac{\partial}{\partial y}u=\frac{\ln x}{z}x^\frac{y}{z}.$$
$$\frac{\partial}{\partial z}u=-\frac{x^\frac{y}{z}y\ln x}{z^2}.$$
Work Step by Step
Regard $y$ and $z$ as constant and differentiate with respect to $x$:
$$\frac{\partial}{\partial x}u=\frac{\partial}{\partial x}x^{y/z}=\frac{y}{z}x^{\frac{y}{z}-1}=\frac{y}{z}x^\frac{y-z}{z}.$$
Regard $x$ and $z$ as constant and differentiate with respect to $y$:
$$\frac{\partial}{\partial y}u=\frac{\partial}{\partial y}x^{y/z}=x^{y/z}\ln x\frac{\partial}{\partial y}\frac{y}{z}=\frac{\ln x}{z}x^\frac{y}{z}.$$
Regard $x$ and $y$ and differentiate with respect to $z$:
$$\frac{\partial}{\partial z}u=\frac{\partial}{\partial z}x^{y/z}=x^{y/z}\ln x\frac{\partial}{\partial z}\frac{y}{z}=x^\frac{y}{z}\ln x\text{ }y\left(-\frac{1}{z^2}\right)=-\frac{x^\frac{y}{z}y\ln x}{z^2}.$$
