While the derivative of $x^3$ with respect to x is
$$\dfrac{d}{dx}x^3 = 3x^2,$$
the derivative of $x^3$ with respect to time t is
$$\dfrac{d}{dt}x^3 = 3x^2\dfrac{dx}{dt}.$$
(Recall that that rate is $\dfrac{dx}{dt} = 6$ in this problem.)

The Chain rule term would be clearer if we explicitly wrote the time-dependence of both y and x. Remember that both quantities change as time passes, and so they are functions of time t. We could have captured this time-dependence explicitly by writing our original relation as
$$ y(t) = \big[ x(t)\big]^3 + 2 \big[ x(t)\big]$$
to remind ourselves that both y and x are functions of time t. Then when we take the derivative,
$$\frac{d}{dt}y(t) = \frac{d}{dt} \big[ x(t)\big]^3 + 2\dfrac{d}{dt} \big[ x(t)\big]$$
Now the term $\dfrac{d}{dt}y(t)$ is the rate $\dfrac{dy(t)}{dt}.$

And looking at the last term, $\dfrac{d}{dt}x(t)$ is the rate $\dfrac{dx(t)}{dt}.$

The derivative $\dfrac{d}{dt}\big[\text{(stuff)} \big]^3 = 3 \text{(stuff)}^2 \cdot \dfrac{d}{dt}\text{(stuff)},$ where the extra term at the end is due to the Chain rule. Here $\text{(stuff)} = x(t)$, so we have $\dfrac{d}{dt}\big[ x(t)\big]^3 = 3\big[ x(t)\big]^2 \dfrac{dx(t)}{dt}.$

Most people find that writing the explicit time-dependence $y(t)$ and x(t) annoying, and so just write y and x instead. Regardless, you must remember that both y and x depend on t, and so when you take the derivative with respect to time the Chain Rule applies and you have the $\dfrac{dy}{dt}$ and $\dfrac{dx}{dt}$ terms.