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 $x$.

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

2. Rearrange the equation to make $\frac{dy}{dx}$ the subject and simplify.

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

1. Substitute the equation of the line into the equation of the curve.

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

2. Simplify the equation to solve for $x$.

$4x^3 - 3x^3 = -x^3 - 2$
$x^3 = -x^3 - 2$
$2x^3 = -2$
$x^3 = -1$
$x = -1$

3. Calculate the corresponding $y$-coordinate and state the coordinates of $P$.

Since $y = -x$,
$y = -(-1) = 1$
Therefore, the coordinates of $P$ are $(-1, 1)$.

1. Compute the derivative and the coordinates of $P$ (required from earlier parts).

From part (a), differentiating $4x^3+3x^2y=y^3-2$ with respect to $x$ gives:
$12x^2 + 6xy + 3x^2 \frac{dy}{dx} = 3y^2 \frac{dy}{dx}$
$\frac{dy}{dx} = \frac{4x^2 + 2xy}{y^2 - x^2}$

From part (b)(i), substituting $y = -x$ into the curve equation to find $P$:
$4x^3 + 3x^2(-x) = (-x)^3 - 2$
$4x^3 - 3x^3 = -x^3 - 2$
$2x^3 = -2 \implies x = -1$
Thus $y = 1$, giving $P(-1, 1)$.

2. Evaluate the gradient of the tangent at $P$.

Substitute $x = -1$ and $y = 1$ into the expression for $\frac{dy}{dx}$:
$\text{Numerator} = 4(-1)^2 + 2(-1)(1) = 4 - 2 = 2$
$\text{Denominator} = (1)^2 - (-1)^2 = 1 - 1 = 0$
Since the denominator is zero while the numerator is non-zero, the value of $\frac{dy}{dx}$ is undefined. This indicates that the tangent is a vertical line.

3. State the equation of the tangent.

A vertical line passing through $P(-1, 1)$ takes the form $x = k$, where $k$ is the $x$-coordinate of the point.
Thus, the equation of the tangent is:
$x = -1$

1. Determine the equation of the normal at $P$ and find the coordinates of $Q$ and $R$.

From earlier parts, $P$ is $(-1, 1)$ and the tangent at $P$ is the vertical line $x = -1$.
Since the normal is perpendicular to the tangent, it must be the horizontal line passing through $P(-1, 1)$, which is $y = 1$.
The tangent $x = -1$ cuts the $x$-axis at $R(-1, 0)$.
The normal $y = 1$ cuts the $y$-axis at $Q(0, 1)$.

2. Deduce the geometrical description of the quadrilateral $OQPR$.

The vertices of the quadrilateral are $O(0, 0)$, $Q(0, 1)$, $P(-1, 1)$, and $R(-1, 0)$.
The sides $OQ$ and $PR$ are vertical with a length of 1 unit.
The sides $QP$ and $RO$ are horizontal with a length of 1 unit.
Since adjacent sides are parallel to the perpendicular $x$ and $y$ axes, all internal angles are $90^\circ$ and all sides are equal in length.
Therefore, $OQPR$ is a square.