Answer
$${\mathbf{G}(x, y,z)=(-\frac{z}{x^2}-\frac{ z}{y} )\mathbf{i}+ (\frac{1}{z} +\frac{x z}{y^2})\mathbf{j}+( -\frac{y}{z^2}+\frac{1}{x}-\frac{x}{y})\mathbf{k}}$$
Work Step by Step
Given
$$g(x, y,z)=\frac{y}{z}+\frac{z}{x}-\frac{x z}{y}$$
Since \begin{align}\mathbf{G}(x,y,z)&=M \mathbf{i}+N\mathbf{j}+P\mathbf{k}\\
&=g_{x}(x, y,z)\mathbf{i}+g_{y}(x, y,z)\mathbf{j}+g_{z}(x, y,z)\mathbf{k}
\end{align}
As, we have
\begin{array}{l}
{g_{x}(x, y,z)=\frac{\partial g(x,y,z)}{\partial x}= -\frac{z}{x^2}-\frac{ z}{y} } \\
{g_{y}(x, y,z)=\frac{\partial g(x,y,z)}{\partial y}= \frac{1}{z} +\frac{x z}{y^2} } \\
{g_{z}(x, y,z)=\frac{\partial g(x,y,z)}{\partial z}=-\frac{y}{z^2}+\frac{1}{x}-\frac{x}{y} } \\ \end{array}
So, we get
$${\mathbf{G}(x, y,z)=(-\frac{z}{x^2}-\frac{ z}{y} )\mathbf{i}+ (\frac{1}{z} +\frac{x z}{y^2})\mathbf{j}+( -\frac{y}{z^2}+\frac{1}{x}-\frac{x}{y})\mathbf{k}}$$
