Supposed that $x$ and $y$ are nonzero real numbers such that $\frac{3x+y}{x-3y}=-2$. What is the value of $\frac{x+3y}{3x-y}$?
$\textbf{(A)}\ -3\qquad\textbf{(B)}\ -1\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ 3$
Rearranging, we find $3x+y=-2x+6y$, or $5x=5y\implies x=y$.
Substituting, we can convert the second equation into $\frac{x+3x}{3x-x}=\frac{4x}{2x}=\boxed{\textbf{(D)}\ 2}$.

More step-by-step explanation:
$\frac{3x+y}{x-3y}=-2$
$3x+y=-2\left(x-3y\right)$
$3x+y=-2x+6y$
$5x=5y$
$x=y$
$\frac{x+3y}{3x-y}=\frac{1+3\left(1\right)}{3\left(1\right)-1}=\frac{4}{2}=2$.
We choose $\boxed{\textbf{(D)}\ 2}$.