To differentiate the function $y = \tan^{-1}(3 \tan x)$, we will apply the chain rule and the derivative of the inverse tangent function.

Step-by-Step Differentiation:

1. Start with the function: $y = \tan^{-1}(3 \tan x)$

2. Differentiate both sides with respect to $x$. For the right-hand side, use the chain rule. The derivative of $\tan^{-1}(u)$ is $\frac{1}{1 + u^2}$, where $u = 3 \tan x$.

   So, applying the chain rule: $\frac{dy}{dx} = \frac{1}{1 + (3 \tan x)^2} \cdot \frac{d}{dx}(3 \tan x)$

3. Differentiate $3 \tan x$: $\frac{d}{dx}(3 \tan x) = 3 \sec^2 x$

4. Substitute this into the previous expression: $\frac{dy}{dx} = \frac{1}{1 + 9 \tan^2 x} \cdot 3 \sec^2 x$

5. Simplify: The final derivative is: $\frac{dy}{dx} = \frac{3 \sec^2 x}{1 + 9 \tan^2 x}$

Thus, the derivative of $y = \tan^{-1}(3 \tan x)$ is:

$\boxed{\frac{dy}{dx} = \frac{3 \sec^2 x}{1 + 9 \tan^2 x}}$

Would you like further details on this, or any clarifications?

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Here are some related questions you could explore:

1. What is the general rule for differentiating inverse trigonometric functions?
2. How do you apply the chain rule in more complex cases?
3. What are the derivatives of the other inverse trigonometric functions?
4. Can we simplify the result further using trigonometric identities?
5. How does the derivative behave as $x$ approaches certain values?

Tip: Always check if simplifying the expression using trigonometric identities can make the differentiation easier.