(i) To find the gradient, we need to differentiate the equation implicitly with respect to \(x\):

\(\frac{d}{dx}(3x^2) - \frac{d}{dx}(4xy) + \frac{d}{dx}(y^2) = \frac{d}{dx}(45)\).

This gives \(6x - (4y + 4x\frac{dy}{dx}) + 2y\frac{dy}{dx} = 0\).

Rearrange to find \(\frac{dy}{dx}\):

\(6x - 4y = (4x - 2y)\frac{dy}{dx}\).

\(\frac{dy}{dx} = \frac{6x - 4y}{4x - 2y}\).

Substitute \(x = 2\) and \(y = -3\):

\(\frac{dy}{dx} = \frac{6(2) - 4(-3)}{4(2) - 2(-3)} = \frac{12 + 12}{8 + 6} = \frac{24}{14} = \frac{12}{7}\).

(ii) To show there are no points where the gradient is 1, set \(\frac{dy}{dx} = 1\):

\(\frac{6x - 4y}{4x - 2y} = 1\).

This implies \(6x - 4y = 4x - 2y\).

Simplifying gives \(2x = 2y\) or \(y = x\).

Substitute \(y = x\) into the original equation:

\(3x^2 - 4x^2 + x^2 = 45\).

This simplifies to \(0 = 45\), a contradiction.

Therefore, there are no points on the curve where the gradient is 1.