A particle moves along the curve $y=\dfrac{5}{(3x-1)^2}$, where $x \ne \dfrac{1}{3}$, in such a way that the x-coordinate is decreasing at a rate of $0.1$ units per second.

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

$y = 5(3x-1)^{-2}$

$\frac{dy}{dx} = 5(-2)(3x-1)^{-3}(3)$
$\frac{dy}{dx} = -30(3x-1)^{-3} = -\frac{30}{(3x-1)^3}$

2. Evaluate $\frac{dy}{dx}$ at $x=1$

When $x = 1$,
$\frac{dy}{dx} = -\frac{30}{(3(1)-1)^3} = -\frac{30}{8} = -3.75$

3. Use connected rates of change to find $\frac{dy}{dt}$

Given $\frac{dx}{dt} = -0.1$ (since $x$ is decreasing),
$\frac{dy}{dt} = \frac{dy}{dx} \times \frac{dx}{dt}$
$\frac{dy}{dt} = -3.75 \times (-0.1) = 0.375$

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

$y = 5(3x-1)^{-2}$
$\frac{dy}{dx} = 5(-2)(3x-1)^{-3}(3) = -30(3x-1)^{-3}$

2. Set up the connected rates of change equation and solve for $x$

Using the chain rule, $\frac{dy}{dt} = \frac{dy}{dx} \times \frac{dx}{dt}$
$\frac{3}{125} = -30(3x-1)^{-3} \times (-0.1)$
$\frac{3}{125} = 3(3x-1)^{-3}$
$\frac{1}{125} = \frac{1}{(3x-1)^3}$
$(3x-1)^3 = 125$
$3x - 1 = 5$
$3x = 6$
$x = 2$