Find the domain of: `y=sin^(-1)((x^(2))/(1+x^(2)))`

Explanation

To find the domain of the function \( y = \sin^{-1}\left(\frac{x^2}{1+x^2}\right) \), we need to ensure that the expression inside the inverse sine function, \( \frac{x^2}{1+x^2} \), falls within the valid range for the sine inverse function.

### Step-by-Step Solution:

1. **Identify the Range of the Inverse Sine Function**:
The function \( \sin^{-1}(t) \) is defined for \( t \) in the interval \([-1, 1]\). Therefore, we need to find the values of \( x \) such that:
\[ -1 \leq \frac{x^2}{1+x^2} \leq 1 \]

2. **Analyze the Expression**:
- The term \( x^2 \) is always non-negative (\( x^2 \geq 0 \)).
- The denominator \( 1 + x^2 \) is always positive (\( 1 + x^2 > 0 \)).
- Therefore, \( \frac{x^2}{1+x^2} \) is always non-negative.

3. **Establish the Lower Bound**:
Since \( \frac{x^2}{1+x^2} \geq 0 \), the lower bound condition \( -1 \leq \frac{x^2}{1+x^2} \) is automatically satisfied for all \( x \).

4. **Establish the Upper Bound**:
Now, we need to check the upper bound:
\[ \frac{x^2}{1+x^2} \leq 1 \]
This inequality can be simplified:
\[ x^2 \leq 1 + x^2 \]
This is always true since \( 1 + x^2 \) is greater than \( x^2 \).

5. **Conclusion**:
Since both conditions for the domain are satisfied for all real numbers \( x \), we conclude that the function \( y = \sin^{-1}\left(\frac{x^2}{1+x^2}\right) \) is defined for all \( x \in \mathbb{R} \).

### Final Answer:
The domain of the function is:
\[ \text{Domain} = \mathbb{R} \quad (\text{all real numbers}) \]