Given that $y = \frac{6x^2}{2-5x}$, find the range of values of $x$ for which $y$ is a decreasing function of $x$.

1. Find the derivative of $y$ with respect to $x$ using the quotient rule.

Let $u = 6x^2$ and $v = 2-5x$.
Then $\frac{du}{dx} = 12x$ and $\frac{dv}{dx} = -5$.
$\frac{dy}{dx} = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}$
$\frac{dy}{dx} = \frac{(2-5x)(12x) - (6x^2)(-5)}{(2-5x)^2}$

2. Simplify the derivative.

$\frac{dy}{dx} = \frac{24x - 60x^2 + 30x^2}{(2-5x)^2}$
$\frac{dy}{dx} = \frac{24x - 30x^2}{(2-5x)^2}$
$\frac{dy}{dx} = \frac{6x(4 - 5x)}{(2-5x)^2}$

3. Set up the condition for a decreasing function.

For $y$ to be a decreasing function, $\frac{dy}{dx} < 0$.
$\frac{6x(4 - 5x)}{(2-5x)^2} < 0$

4. Solve the inequality.

Since $(2-5x)^2 > 0$ for all $x \neq \frac{2}{5}$, the sign of the derivative depends only on the numerator.
$6x(4 - 5x) < 0$
The roots of $6x(4 - 5x) = 0$ are $x = 0$ and $x = \frac{4}{5}$.
The quadratic expression represents a parabola opening downwards (since the coefficient of $x^2$ is $-30$).
Therefore, the expression is negative when $x < 0$ or $x > \frac{4}{5}$.