(i) Express \(f(x)\) in partial fractions:

Assume \(\frac{3x}{(1+x)(1+2x^2)} = \frac{A}{1+x} + \frac{Bx+C}{1+2x^2}\).

Multiply through by \((1+x)(1+2x^2)\) to clear the denominators:

\(3x = A(1+2x^2) + (Bx+C)(1+x)\).

Expand and equate coefficients:

\(3x = A + 2Ax^2 + Bx + Bx^2 + C + Cx\).

Equating coefficients gives:

\(A + C = 0\), \(B + C = 3\), \(2A + B = 0\).

Solving these equations, we find \(A = -1\), \(B = 2\), \(C = 1\).

Thus, \(f(x) = \frac{-1}{1+x} + \frac{2x+1}{1+2x^2}\).

(ii) Expand \(f(x)\) in ascending powers of \(x\):

First, expand \(\frac{-1}{1+x}\) using the geometric series:

\(\frac{-1}{1+x} = -(1 - x + x^2 - x^3 + \cdots)\).

Next, expand \(\frac{2x+1}{1+2x^2}\):

\(\frac{2x+1}{1+2x^2} = (2x+1)(1 - 2x^2 + 4x^4 - \cdots)\).

Multiply out to get terms up to \(x^3\):

\((2x+1)(1 - 2x^2) = 2x + 1 - 4x^3 - 2x^2\).

Combine the expansions:

\(f(x) = -(1 - x + x^2 - x^3) + (2x + 1 - 2x^2 - 4x^3)\).

Simplify to get:

\(f(x) = 3x - 3x^2 - 3x^3\).