Suppose that $x$ and $y$ are non-zero real numbers such that $\frac{x}{3} = y^2$ and $\frac{x}{9} = 9y$. Find the value of $x - y$.

1. Express $x$ in terms of $y$ using the linear equation

From $\frac{x}{9} = 9y$, we have:
$x = 81y$

2. Substitute into the other equation to solve for $y$

Substitute $x = 81y$ into $\frac{x}{3} = y^2$:
$\frac{81y}{3} = y^2$
$27y = y^2$
$y^2 - 27y = 0$
$y(y - 27) = 0$
Since $y$ is a non-zero real number, $y = 27$.

3. Find $x$ and calculate $x - y$

$x = 81(27) = 2187$
$x - y = 2187 - 27 = 2160$