To solve the equations, we need to clear the fractions by finding a common denominator and then simplify.

### Equation (i):
\[ \frac{x+3}{x-3} = \frac{3}{x^2-3x} + \frac{1}{x} \]

1. Combine the terms on the right side:
\[
\frac{3}{x(x-3)} + \frac{1}{x} = \frac{3 + (x-3)}{x(x-3)} = \frac{x}{x(x-3)}
\]

2. Simplify:
\[
\frac{x+3}{x-3} = \frac{x}{x(x-3)}
\]

3. Cross-multiply:
\[
(x+3)x = x(x-3)
\]

4. Simplify:
\[
x^2 + 3x = x^2 - 3x
\]

5. Solve:
\[
6x = 0 \Rightarrow x = 0
\]

Check for restrictions (denominators cannot be zero). Here, \(x \neq 3\) and \(x \neq 0\), so no solution exists for this equation when considering the domain.

### Equation (h):
\[ \frac{2x+1}{x+1} + \frac{2}{x} = \frac{2}{x(x+1)} \]

1. Combine the left side over a common denominator \(x(x+1)\):
\[
\frac{(2x+1)x + 2(x+1)}{x(x+1)} = \frac{2x+1+2x+2}{x(x+1)}
\]

2. Simplify:
\[
\frac{4x+3}{x(x+1)} = \frac{2}{x(x+1)}
\]

3. Cross-multiply:
\[
(4x+3) = 2
\]

4. Solve:
\[
4x = -1 \Rightarrow x = -\frac{1}{4}
\]

Check for restrictions: \(x \neq -1\) and \(x \neq 0\).

### Equation (j):
\[ \frac{1}{x} - \frac{x+4}{x-4} = \frac{4}{4x-x^2} \]

1. Simplify right side:
\[
\frac{4}{x(4-x)} = \frac{4}{-x(x-4)} = -\frac{4}{x(x-4)}
\]

2. Combine left side over a common denominator \(x(x-4)\):
\[
\frac{x-4-(x+4)}{x(x-4)} = \frac{-8}{x(x-4)}
\]

3. Simplify:
\[
-\frac{8}{x(x-4)} = -\frac{4}{x(x-4)}
\]

4. Cross-multiply:
\[
-8 = -4
\]

This will not yield a valid result; the process suggests there's no solution.

Check for restrictions: \(x \neq 0\) and \(x \neq 4\).

**Conclusion:**
- Equation (i) has no solution.
- Equation (h) solves to \(x = -\frac{1}{4}\).
- Equation (j) has no solution.