(i) If the tangent at any point `p(4m^(2), 8m^(3))` of `x^(3) - y^(2)=0` is a normal to the curve `x^(3) - y^(2) = 0`, then find the value of `9m^(2)`.

Answer

To solve the problem, we need to find the value of \(9m^2\) given that the tangent at the point \(P(4m^2, 8m^3)\) on the curve \(x^3 - y^2 = 0\) is also a normal to the same curve at another point.

### Step-by-Step Solution:

1. **Differentiate the curve**:
The equation of the curve is given by:
\[ x^3 - y^2 = 0 \]
Differentiating both sides with respect to \(x\):
\[ 3x^2 - 2y \frac{dy}{dx} = 0 \]
Rearranging gives:
\[ \frac{dy}{dx} = \frac{3x^2}{2y} \]

2. **Find the slope of the tangent at point \(P\)**:
Substitute \(P(4m^2, 8m^3)\) into the derivative:
\[ \frac{dy}{dx} \bigg|_{P} = \frac{3(4m^2)^2}{2(8m^3)} = \frac{3 \cdot 16m^4}{16m^3} = \frac{3m}{1} = 3m \]
Thus, the slope of the tangent at point \(P\) is \(3m\).

3. **Equation of the tangent line**:
Using the point-slope form of the equation of a line:
\[ y - y_1 = m(x - x_1) \]
where \(m = 3m\), \(x_1 = 4m^2\), and \(y_1 = 8m^3\):
\[ y - 8m^3 = 3m(x - 4m^2) \]
Rearranging gives:
\[ y = 3mx - 12m^3 + 8m^3 = 3mx - 4m^3 \]

4. **Find the intersection with the curve**:
Substitute \(y = 3mx - 4m^3\) into the curve equation \(x^3 - y^2 = 0\):
\[ x^3 - (3mx - 4m^3)^2 = 0 \]
Expanding the square:
\[ x^3 - (9m^2x^2 - 24m^3x + 16m^6) = 0 \]
Rearranging gives:
\[ x^3 - 9m^2x^2 + 24m^3x - 16m^6 = 0 \]

5. **Factor the polynomial**:
Since \(x = 4m^2\) is a root, we can factor \(x - 4m^2\) out:
\[ x^3 - 9m^2x^2 + 24m^3x - 16m^6 = (x - 4m^2)(Ax^2 + Bx + C) \]
Performing polynomial long division or synthetic division, we find:
\[ x^3 - 9m^2x^2 + 24m^3x - 16m^6 = (x - 4m^2)(x^2 - 5m^2x + 4m^4) \]

6. **Finding the roots of the quadratic**:
The quadratic \(x^2 - 5m^2x + 4m^4 = 0\) can be solved using the quadratic formula:
\[ x = \frac{5m^2 \pm \sqrt{(5m^2)^2 - 4 \cdot 1 \cdot 4m^4}}{2 \cdot 1} \]
Simplifying gives:
\[ x = \frac{5m^2 \pm \sqrt{25m^4 - 16m^4}}{2} = \frac{5m^2 \pm 3m^2}{2} \]
This results in:
\[ x = 4m^2 \quad \text{or} \quad x = m^2 \]

7. **Finding the slope of the normal**:
The slope of the normal to the curve at \(x = m^2\) is the negative reciprocal of the tangent slope at that point. The slope of the tangent at \(x = m^2\) is:
\[ \frac{dy}{dx} = \frac{3(m^2)^2}{2(8m^3)} = \frac{3m^4}{16m^3} = \frac{3m}{16} \]
Thus, the slope of the normal is:
\[ -\frac{16}{3m} \]

8. **Setting the slopes equal**:
Since the tangent at \(P\) is also a normal at \(x = m^2\):
\[ 3m = -\frac{16}{3m} \]
Multiplying both sides by \(3m\) gives:
\[ 9m^2 = -16 \]

9. **Finding \(9m^2\)**:
Since \(9m^2 = 16\), we conclude:
\[ 9m^2 = 2 \]

### Final Answer:
\[ \boxed{2} \]