If `f = (x)/(1+x^(2)) + (1)/(3) ((x)/(1+x^(2)))^(3) + (1)/(5) ((x)/(1+x^(2)))^(5) + ...`
and `g = x- (2)/(3)x^(3) + (1)/(5)x^(5) + (1)/(7) x^(7) - (2)/(9)x^(9) +...` then `f -=g`.

Answer

To solve the problem, we need to analyze the functions \( f \) and \( g \) given in the question and then find \( f - g \).

### Step 1: Express \( f \) in terms of a known series
The function \( f \) is given as:
\[ f = \frac{x}{1+x^2} + \frac{1}{3} \left( \frac{x}{1+x^2} \right)^3 + \frac{1}{5} \left( \frac{x}{1+x^2} \right)^5 + \ldots \]

This series can be recognized as a Taylor series expansion. We can relate it to the series for \( \log \) functions. The series for \( \log(1+x) \) can be expressed as:
\[ \log(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \ldots \]

### Step 2: Identify the series form
Notice that the series for \( f \) can be rewritten using the substitution \( u = \frac{x}{1+x^2} \):
\[ f = \sum_{n=0}^{\infty} \frac{1}{2n+1} u^{2n+1} = \frac{1}{2} \log \left( \frac{1+u}{1-u} \right) \]

### Step 3: Simplify \( f \)
Substituting \( u \):
\[ f = \frac{1}{2} \log \left( \frac{1 + \frac{x}{1+x^2}}{1 - \frac{x}{1+x^2}} \right) \]
This simplifies to:
\[ f = \frac{1}{2} \log \left( \frac{(1+x^2+x)}{(1+x^2-x)} \right) \]

### Step 4: Express \( g \)
The function \( g \) is given as:
\[ g = x - \frac{2}{3} x^3 + \frac{1}{5} x^5 + \frac{1}{7} x^7 - \frac{2}{9} x^9 + \ldots \]

This series can also be related to the logarithmic series. We can rewrite \( g \) as:
\[ g = \sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1} x^{2n+1} - \frac{2}{3} x^3 - \frac{2}{9} x^9 \]

### Step 5: Combine \( f \) and \( g \)
Now we need to find \( f - g \):
\[ f - g = \left( \frac{1}{2} \log \left( \frac{(1+x^2+x)}{(1+x^2-x)} \right) \right) - \left( x - \frac{2}{3} x^3 + \frac{1}{5} x^5 + \ldots \right) \]

### Step 6: Analyze the result
To analyze the result, we can evaluate \( f \) and \( g \) at specific points, such as \( x = 0 \) and \( x = 1 \).

1. At \( x = 0 \):
\[ f(0) = 0, \quad g(0) = 0 \quad \Rightarrow \quad f(0) - g(0) = 0 \]

2. At \( x = 1 \):
\[ f(1) = \frac{1}{2} \log(3), \quad g(1) \text{ is undefined.} \]

### Conclusion
Thus, we find that \( f \) and \( g \) are not equal for all \( x \), specifically at \( x = 1 \). Therefore, the statement \( f - g = 0 \) is false.