The equation of a curve is given by $4x^3+3x^2y=y^3-2$.

1. Differentiate both sides of the given equation implicitly with respect to $x$, applying the product rule and chain rule.

$\frac{d}{dx}(4x^3) + \frac{d}{dx}(3x^2y) = \frac{d}{dx}(y^3-2)$
$12x^2 + \left(6xy + 3x^2\frac{dy}{dx}\right) = 3y^2\frac{dy}{dx}$

2. Collect all terms containing $\frac{dy}{dx}$ on one side to isolate the derivative.

$12x^2 + 6xy = 3y^2\frac{dy}{dx} - 3x^2\frac{dy}{dx}$
$12x^2 + 6xy = \frac{dy}{dx}(3y^2 - 3x^2)$

3. Make $\frac{dy}{dx}$ the subject and simplify the expression by extracting and cancelling the common factor.

$\frac{dy}{dx} = \frac{12x^2 + 6xy}{3y^2 - 3x^2}$
$\frac{dy}{dx} = \frac{3(4x^2 + 2xy)}{3(y^2 - x^2)}$
$\frac{dy}{dx} = \frac{4x^2 + 2xy}{y^2 - x^2}$

1. Substitute the equation of the line into the equation of the curve and solve for $x$.

Substitute $y = -x$ into $4x^3+3x^2y=y^3-2$:
$4x^3 + 3x^2(-x) = (-x)^3 - 2$
$4x^3 - 3x^3 = -x^3 - 2$
$x^3 = -x^3 - 2$
$2x^3 = -2$
$x^3 = -1$
$x = -1$

2. Find the corresponding $y$-coordinate to obtain the coordinates of $P$.

Substitute $x = -1$ into $y = -x$:
$y = -(-1) = 1$

Therefore, the coordinates of $P$ are $(-1, 1)$.

1. Find the coordinates of $P$ (from part b(i)).

Substitute $y = -x$ into the curve equation $4x^3+3x^2y=y^3-2$:
$4x^3 + 3x^2(-x) = (-x)^3 - 2$
$x^3 = -x^3 - 2 \implies 2x^3 = -2 \implies x = -1$
Since $y = -x$, we have $y = 1$. Thus, $P$ is $(-1, 1)$.

2. Find the expression for $\frac{dy}{dx}$ (from part a).

Differentiating the curve equation implicitly with respect to $x$:
$\frac{d}{dx}(4x^3 + 3x^2y) = \frac{d}{dx}(y^3 - 2)$
$12x^2 + 6xy + 3x^2\frac{dy}{dx} = 3y^2\frac{dy}{dx}$
$\frac{dy}{dx} = \frac{12x^2 + 6xy}{3y^2 - 3x^2} = \frac{4x^2 + 2xy}{y^2 - x^2}$

3. Evaluate the gradient of the tangent at point $P$.

Substitute $x = -1$ and $y = 1$ into $\frac{dy}{dx}$:
$\frac{dy}{dx} = \frac{4(-1)^2 + 2(-1)(1)}{(1)^2 - (-1)^2} = \frac{4 - 2}{1 - 1} = \frac{2}{0}$
The denominator is $0$ and the numerator is non-zero, so the gradient is undefined. This indicates that the tangent to the curve is a vertical line.

4. Determine the equation of the tangent at $P$.

A vertical tangent passing through $P(-1, 1)$ has a constant $x$-coordinate.
The equation of the tangent is therefore $x = -1$.

1. Find the coordinates of $P$ and the equations of the tangent and normal at $P$ (from part b).

Substitute $y = -x$ into the curve equation $4x^3 + 3x^2y = y^3 - 2$:
$4x^3 - 3x^3 = -x^3 - 2 \implies 2x^3 = -2 \implies x = -1$
Hence $y = 1$, so $P$ is $(-1, 1)$.
From (a), $\frac{dy}{dx} = \frac{4x^2+2xy}{y^2-x^2}$. At $P(-1,1)$:
$\frac{dy}{dx} \to \frac{4(1) - 2}{1 - 1} = \frac{2}{0} \text{ (undefined)}$
Thus, the tangent at $P$ is the vertical line $x = -1$, and the normal at $P$ is the horizontal line $y = 1$.

2. Determine the coordinates of $Q$ and $R$.

The vertical tangent $x = -1$ intersects the x-axis at $R(-1, 0)$.
The horizontal normal $y = 1$ intersects the y-axis at $Q(0, 1)$.

3. Describe the geometrical shape of the quadrilateral $OQPR$.

The vertices of the quadrilateral are $O(0,0)$, $Q(0,1)$, $P(-1,1)$, and $R(-1,0)$.
Since $OQ$ lies on the y-axis and $OR$ lies on the x-axis, $\angle ROQ = 90^\circ$.
The lengths of the sides are:
$OQ = 1$
$QP = |-1 - 0| = 1$
$PR = |1 - 0| = 1$
$RO = |-1 - 0| = 1$
Because all four sides are of equal length and the internal angles are $90^\circ$, $OQPR$ is a square.