If $x+y=12$ and $xy=6$, then the value of $\frac{x}{y}+\frac{y}{x}$ is equal to:

We need to find the value of:
$$\frac{x}{y}+\frac{y}{x}=\frac{x^2+y^2}{xy}$$
We know the algebraic identity:
$$x^2+y^2=(x+y{)}^2-2xy$$
Given $x+y=12$ and $xy=6$, substitute these values:
$$x^2+y^2=12^2-2(6)=144-12=132$$
Now, substitute this back into the expression:
$$\frac{x^2+y^2}{xy}=\frac{132}{6}=22$$