`y=tan^(-1)""(3x-x^(3))/(2x^(2)-1),-(1)/(sqrt(3))ltxlt(1)/(sqrt(3))`

Explanation

To solve the problem step by step, we will differentiate the function \( y = \tan^{-1} \left( \frac{3x - x^3}{2x^2 - 1} \right) \).

### Step 1: Rewrite the function
Let:
\[ y = \tan^{-1} \left( \frac{3x - x^3}{2x^2 - 1} \right) \]

### Step 2: Use the substitution
We will use the substitution \( x = \tan(\theta) \). Therefore, we have:
\[ \frac{3x - x^3}{2x^2 - 1} = \frac{3\tan(\theta) - \tan^3(\theta)}{2\tan^2(\theta) - 1} \]

### Step 3: Recognize the tangent triple angle formula
The expression \( \frac{3\tan(\theta) - \tan^3(\theta)}{2\tan^2(\theta) - 1} \) can be simplified using the tangent triple angle formula:
\[ \tan(3\theta) = \frac{3\tan(\theta) - \tan^3(\theta)}{1 - 3\tan^2(\theta)} \]
Thus, we can rewrite:
\[ y = \tan^{-1}(\tan(3\theta)) \]

### Step 4: Simplify using the properties of inverse tangent
Since \( y = \tan^{-1}(\tan(3\theta)) \), we have:
\[ y = 3\theta \]
This is valid as long as \( 3\theta \) is within the range of the inverse tangent function.

### Step 5: Substitute back for \( \theta \)
Recall that \( x = \tan(\theta) \), so:
\[ \theta = \tan^{-1}(x) \]
Thus:
\[ y = 3\tan^{-1}(x) \]

### Step 6: Differentiate with respect to \( x \)
Now, we differentiate \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = 3 \cdot \frac{d}{dx}(\tan^{-1}(x)) \]
We know that:
\[ \frac{d}{dx}(\tan^{-1}(x)) = \frac{1}{1 + x^2} \]
So:
\[ \frac{dy}{dx} = 3 \cdot \frac{1}{1 + x^2} \]

### Final Answer
Thus, the derivative is:
\[ \frac{dy}{dx} = \frac{3}{1 + x^2} \]