Express the curve $C$ with parametrization $\{(\cos 3t, \sin 2t) : t \in \mathbb{R} \}$ as a function of $X$ and $Y$.
Here's what I have (I don't know if this is helpful)
I let $X= \cos 3t$ and $Y= \sin 2t$. So I have $X= \cos^3 t -3\sin^2 t \cos t$ and $Y=2 \cos t \sin t$.
I don't know how to go on from here. Please help. Thanks!
Answer: $$ X^4 - X^2 + 4 Y^6 - 6 Y^4 + \frac{9}{4} Y^2 = 0 $$
How I got it:
We can replace the trigonometric parametrization $(\cos(u), \sin(u))$ of the unit circle with a rational one: $\left(\frac{1-t^2}{1+t^2}, \frac{2t}{1+t^2}\right)$. So we equate $\cos(u) = x = \frac{1-t^2}{1+t^2}$ and $\sin(u) = y =\frac{1-t^2}{1+t^2}$. Your expressions for $X$ and $Y$ then become \begin{align*} X &= \cos^3(u) - 3 \cos(u) \sin^2(u) = x^3 - 3 xy^2 = \frac{-t^{6} + 15 t^{4} - 15 t^{2} + 1}{(1+t^2)^3}\\ Y &= 2 \cos(u) \sin(u) = 2 x y = \frac{-4 t^{3} + 4 t}{(1+t^2)^2} \end{align*}
In the polynomial ring $\mathbb{Q}[t,z,X,Y]$ we form the ideal $$ I = ((1+t^2)^3 X -(-t^{6} + 15 t^{4} - 15 t^{2} + 1), (1+t^2)^2 Y - (-4 t^{3} + 4 t), (1+t^2)z - 1) $$ and compute a Gröbner basis with respect to the elimination ordering $t > z > X > Y$. The last entry of this Gröbner basis is the desired polynomial in $X$ and $Y$.
The proof is in the pudding, though: here is a SageMathCell comparing the two plots.
Actually, a slightly better way is to instead compute a Gröbner basis for the ideal $(X - (x^3 - 3xy^2), Y - (2xy), x^2 + y^2 - 1)$ in $\mathbb{Q}[x,y,X,Y]$ with respect to the elimination ordering $x > y > X > Y$. This yields the same answer.