Let `f : R-{1} to R` be defined by
`f(x) =(1+x)/(1-x) AA x in R -{1}` then for `x ne +-1, (1+f(x)^(2))/(f(x)f(x^(2)))`=………….

Answer

To solve the given problem, we need to evaluate the expression \((1 + f(x)^2) / (f(x) f(x^2))\) where \(f(x) = \frac{1+x}{1-x}\) for \(x \in \mathbb{R} - \{1\}\).

### Step 1: Calculate \(f(x)\)
Given:
\[ f(x) = \frac{1+x}{1-x} \]

### Step 2: Calculate \(f(x^2)\)
Now, we need to find \(f(x^2)\):
\[ f(x^2) = \frac{1+x^2}{1-x^2} \]

### Step 3: Calculate \(f(x)^2\)
Next, we calculate \(f(x)^2\):
\[ f(x)^2 = \left(\frac{1+x}{1-x}\right)^2 = \frac{(1+x)^2}{(1-x)^2} = \frac{1 + 2x + x^2}{1 - 2x + x^2} \]

### Step 4: Substitute \(f(x)\) and \(f(x^2)\) into the expression
Now we substitute \(f(x)\) and \(f(x^2)\) into the expression:
\[ \frac{1 + f(x)^2}{f(x) f(x^2)} = \frac{1 + \frac{1 + 2x + x^2}{1 - 2x + x^2}}{\frac{1+x}{1-x} \cdot \frac{1+x^2}{1-x^2}} \]

### Step 5: Simplify the numerator
The numerator becomes:
\[ 1 + f(x)^2 = 1 + \frac{1 + 2x + x^2}{1 - 2x + x^2} = \frac{(1 - 2x + x^2) + (1 + 2x + x^2)}{1 - 2x + x^2} = \frac{2 + 2x^2}{1 - 2x + x^2} \]

### Step 6: Simplify the denominator
The denominator is:
\[ f(x) f(x^2) = \frac{(1+x)(1+x^2)}{(1-x)(1-x^2)} = \frac{(1+x+x^2+xy)}{(1-x)(1-x^2)} \]

### Step 7: Combine the numerator and denominator
Now we have:
\[ \frac{\frac{2 + 2x^2}{1 - 2x + x^2}}{\frac{(1+x)(1+x^2)}{(1-x)(1-x^2)}} \]

### Step 8: Simplify the entire expression
This simplifies to:
\[ \frac{(2 + 2x^2)(1-x)(1-x^2)}{(1-2x+x^2)(1+x)(1+x^2)} \]

### Step 9: Final simplification
After simplifying, we find that the terms cancel out, leading us to:
\[ = 2 \]

Thus, the final result is:
\[ \frac{(1 + f(x)^2)}{(f(x) f(x^2))} = 2 \]

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