$\theta$ elimination from the equation $x^2 + y^2 = \dfrac{x\cos 3\theta + y\sin 3\theta}{\cos^3\theta} = \dfrac{y\cos 3\theta - x\sin 3\theta}{\sin^3\theta}$ will be

Let $x^2 + y^2 = k$.

From the given condition, we can write the following two equations:
$x \cos 3\theta + y \sin 3\theta = k \cos^3 \theta$
$-x \sin 3\theta + y \cos 3\theta = k \sin^3 \theta$

Solving for $x$ and $y$ by multiplying the first equation by $\cos 3\theta$ and the second by $-\sin 3\theta$, then adding them:
$x(\cos^2 3\theta + \sin^2 3\theta) = k(\cos^3 \theta \cos 3\theta - \sin^3 \theta \sin 3\theta)$
$x = k(\cos^3 \theta \cos 3\theta - \sin^3 \theta \sin 3\theta)$

Similarly, multiplying the first equation by $\sin 3\theta$ and the second by $\cos 3\theta$, then adding them:
$y(\sin^2 3\theta + \cos^2 3\theta) = k(\cos^3 \theta \sin 3\theta + \sin^3 \theta \cos 3\theta)$
$y = k(\cos^3 \theta \sin 3\theta + \sin^3 \theta \cos 3\theta)$

Using the multiple angle formulas $\cos^3 \theta = \dfrac{\cos 3\theta + 3\cos \theta}{4}$ and $\sin^3 \theta = \dfrac{3\sin \theta - \sin 3\theta}{4}$, we substitute these into the expression for $x$:
$x = \dfrac{k}{4} [(\cos 3\theta + 3\cos \theta)\cos 3\theta - (3\sin \theta - \sin 3\theta)\sin 3\theta]$
$x = \dfrac{k}{4} [\cos^2 3\theta + 3\cos \theta \cos 3\theta - 3\sin \theta \sin 3\theta + \sin^2 3\theta]$
$x = \dfrac{k}{4} [1 + 3(\cos 3\theta \cos \theta - \sin 3\theta \sin \theta)]$
$x = \dfrac{k}{4} [1 + 3\cos 4\theta]$

Substituting into the expression for $y$:
$y = \dfrac{k}{4} [(\cos 3\theta + 3\cos \theta)\sin 3\theta + (3\sin \theta - \sin 3\theta)\cos 3\theta]$
$y = \dfrac{k}{4} [\cos 3\theta \sin 3\theta + 3\cos \theta \sin 3\theta + 3\sin \theta \cos 3\theta - \sin 3\theta \cos 3\theta]$
$y = \dfrac{k}{4} [3(\sin 3\theta \cos \theta + \cos 3\theta \sin \theta)]$
$y = \dfrac{k}{4} [3\sin 4\theta]$

From the simplified expressions for $x$ and $y$, we isolate the trigonometric terms containing $4\theta$:
$3\cos 4\theta = \dfrac{4x}{k} - 1 = \dfrac{4x - k}{k}$
$3\sin 4\theta = \dfrac{4y}{k}$

Squaring and adding these two equations to eliminate $\theta$:
$(3\cos 4\theta)^2 + (3\sin 4\theta)^2 = \left(\dfrac{4x - k}{k}\right)^2 + \left(\dfrac{4y}{k}\right)^2$
$9 = \dfrac{16x^2 - 8kx + k^2 + 16y^2}{k^2}$
$9k^2 = 16(x^2 + y^2) - 8kx + k^2$

Substitute back $x^2 + y^2 = k$:
$9k^2 = 16k - 8kx + k^2$
$8k^2 = 16k - 8kx$

Dividing by $8k$ (assuming $k \neq 0$):
$k = 2 - x$
$x^2 + y^2 + x - 2 = 0$

Evaluating option (2) which is $(x^2 + y^2 + 2x)(x^2 + y^2 - x) = 2y^2$:
Substituting $x^2 + y^2 = k$ gives $(k + 2x)(k - x) = 2y^2$
$k^2 + kx - 2x^2 = 2y^2$
$k^2 + kx = 2(x^2 + y^2) = 2k$
$k(k + x - 2) = 0$
Since $k \neq 0$, this simplifies exactly to $x^2 + y^2 + x - 2 = 0$.

Answer: $(x^2 + y^2 + 2x)(x^2 + y^2 - x) = 2y^2$