(i) The gradient of the tangent at \(P\) is the same as the gradient of the curve at \(P\). Since \(TN = 2\), the horizontal distance from \(T\) to \(N\) is 2. The gradient of the tangent is \(\frac{y}{2}\), so \(\frac{dy}{dx} = \frac{1}{2}y\).

(ii) To solve the differential equation \(\frac{dy}{dx} = \frac{1}{2}y\), separate variables: \(\frac{1}{y} dy = \frac{1}{2} dx\).

Integrate both sides: \(\int \frac{1}{y} dy = \int \frac{1}{2} dx\).

This gives \(\ln y = \frac{1}{2}x + C\).

Exponentiate both sides: \(y = e^{\frac{1}{2}x + C} = e^C e^{\frac{1}{2}x}\).

Let \(e^C = A\), so \(y = Ae^{\frac{1}{2}x}\).

Using the point \((4, 3)\), substitute to find \(A\): \(3 = Ae^{2}\).

Thus, \(A = \frac{3}{e^2}\).

The equation of the curve is \(y = \frac{3}{e^2} e^{\frac{1}{2}x} = 3e^{\frac{1}{2}x - 2}\).