A curve $C$ has parametric equations $x = 3t^3,\ y = \frac{3}{1+t^2},\quad t \in \mathbb{R}.$

1. Find the derivatives of $x$ and $y$ with respect to $t$.

$\frac{\text{d}x}{\text{d}t} = 9t^2$

$\frac{\text{d}y}{\text{d}t} = -3(1+t^2)^{-2}(2t) = -\frac{6t}{(1+t^2)^2}$

2. Find $\frac{\text{d}y}{\text{d}x}$ using the parametric chain rule.

$\frac{\text{d}y}{\text{d}x} = \frac{\frac{\text{d}y}{\text{d}t}}{\frac{\text{d}x}{\text{d}t}} = \frac{-\frac{6t}{(1+t^2)^2}}{9t^2}$

$\frac{\text{d}y}{\text{d}x} = -\frac{2}{3t(1+t^2)^2}$

3. Explain why the curve has no stationary points based on $\frac{\text{d}y}{\text{d}x}$.

For $C$ to have a stationary point, we require $\frac{\text{d}y}{\text{d}x} = 0$.

Since the numerator of $\frac{\text{d}y}{\text{d}x}$ is a non-zero constant ($-2$), the fraction can never be equal to zero.

Furthermore, at $t=0$, $\frac{\text{d}y}{\text{d}x}$ is undefined (indicating a vertical tangent, not a stationary point).

Therefore, $\frac{\text{d}y}{\text{d}x} \neq 0$ for all valid $t$, which means $C$ has no stationary points.

1. Determine the intercepts of the curve.

Substitute $t = 0$ to find the $y$-intercept:
$x = 3(0)^3 = 0$
$y = \frac{3}{1+0^2} = 3$
The curve intersects the $y$-axis at $(0, 3)$.

2. Find the equation of the asymptote by considering the behaviour as $t \to \pm\infty$.

As $t \to \infty$, $x \to \infty$ and $y \to 0$.
As $t \to -\infty$, $x \to -\infty$ and $y \to 0$.
Therefore, the horizontal asymptote is the $x$-axis, given by the equation $y = 0$.

3. Sketch the curve, noting the cusp at the $y$-intercept.

Using a graphing calculator or by analyzing the derivative from part (a) ($\frac{\text{d}y}{\text{d}x} \to \pm\infty$ as $t \to 0$), we observe a cusp at the intercept $(0, 3)$.
Sketch a bell-like curve that is symmetric about the $y$-axis, featuring a sharp cusp pointing downwards at $(0, 3)$ and flattening out along the $x$-axis ($y = 0$) in both the positive and negative directions. Label the intercept $(0,3)$ and the asymptote $y=0$.