Step 1: Find the t-derivatives:
Differentiate both given equations with respect to $t$.
\[ \frac{dx}{dt} = 3(\cos t - \cos 3t), \qquad \frac{dy}{dt} = 3(\sin 3t - \sin t) \]
Step 2: Use sum-to-product identities:
We know $\cos C - \cos D = -2\sin\frac{C+D}{2}\sin\frac{C-D}{2}$ and $\sin C - \sin D = 2\cos\frac{C+D}{2}\sin\frac{C-D}{2}$.
With $C = t$, $D = 3t$ the first gives $\cos t - \cos 3t = -2\sin 2t\,\sin(-t) = 2\sin 2t\sin t$.
With $C = 3t$, $D = t$ the second gives $\sin 3t - \sin t = 2\cos 2t\,\sin t$.
\[ \frac{dx}{dt} = 6\sin 2t\,\sin t, \qquad \frac{dy}{dt} = 6\cos 2t\,\sin t \]
Step 3: Divide to get dy/dx:
The common factor $6\sin t$ cancels.
\[ \frac{dy}{dx} = \frac{6\cos 2t\,\sin t}{6\sin 2t\,\sin t} = \cot 2t \]
Step 4: Second derivative by the chain rule:
Since $\dfrac{dy}{dx}$ is a function of $t$, we write $\dfrac{d^2y}{dx^2} = \dfrac{d(\cot 2t)/dt}{dx/dt}$.
The numerator is $-2\operatorname{cosec}^2 2t$ and the denominator is $6\sin 2t\,\sin t$.
\[ \frac{d^2y}{dx^2} = \frac{-2\operatorname{cosec}^2 2t}{6\sin 2t\,\sin t} \]
Step 5: Rewrite with cosec:
Use $\dfrac{1}{\sin 2t} = \operatorname{cosec} 2t$ and $\dfrac{1}{\sin t} = \operatorname{cosec} t$.
\[ \frac{d^2y}{dx^2} = -\frac{1}{3}\operatorname{cosec}^2 2t\cdot\operatorname{cosec} 2t\cdot\operatorname{cosec} t = \frac{-\operatorname{cosec}^3 2t\cdot\operatorname{cosec} t}{3} \]
Final Answer:
We get $\dfrac{dy}{dx} = \cot 2t$ and the stated form of $\dfrac{d^2y}{dx^2}$. \[ \boxed{\frac{dy}{dx} = \cot 2t} \]