Evaluate the following expression for$x=-1,y=-2,z=3$

$\frac{x}{y}+\frac{y}{z}+\frac{z}{x}$

We have to find the value of $\frac{x}{y}+\frac{y}{z}+\frac{z}{x}$

We have $x=-1,y=-2,z=3$

So, replacing $x$ by$-1$, $y$ by$-2$, and $z$ by$3$, We have

$$\begin{gathered} \frac{x}{y}+\frac{y}{z}+\frac{z}{x}=\left(\frac{-1}{-2}\right)+\left(\frac{-2}{3}\right)+\left(\frac{3}{-1}\right) \\ =\left(\frac{1}{2}\right)+\left(\frac{-2}{3}\right)+\left(\frac{-3}{1}\right) \\ =\left(\frac{1\times 3}{2\times 3}\right)+\left(\frac{-2\times 2}{3\times 2}\right)+\left(\frac{-3\times 6}{1\times 6}\right) \\ =\frac{3}{6}+\left(\frac{-4}{6}\right)+\left(\frac{-18}{6}\right) \\ =\frac{3-4-18}{6} \\ =\frac{3-22}{6}=\frac{-19}{6} \end{gathered}$$

Hence, the required value is $\frac{-19}{6}$