The given function is:

$y = \frac{x^2}{x^2 + 1}, \quad x = \sqrt{3t + 1}$

We are tasked with finding $\frac{dy}{dt}$ using the chain rule.

Step-by-step process:

1. Find $\frac{dy}{dx}$: Differentiate $y = \frac{x^2}{x^2 + 1}$ with respect to $x$.

   Using the quotient rule:

   $\frac{dy}{dx} = \frac{(x^2 + 1) \cdot 2x - x^2 \cdot 2x}{(x^2 + 1)^2}$ Simplify the numerator: $\frac{dy}{dx} = \frac{2x(x^2 + 1 - x^2)}{(x^2 + 1)^2} = \frac{2x}{(x^2 + 1)^2}$

2. Find $\frac{dx}{dt}$: Now, differentiate $x = \sqrt{3t + 1}$ with respect to $t$.

   $\frac{dx}{dt} = \frac{1}{2} \cdot (3t + 1)^{-\frac{1}{2}} \cdot 3 = \frac{3}{2\sqrt{3t + 1}}$

3. Apply the chain rule: Now, use the chain rule to find $\frac{dy}{dt}$:

   $\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt}$ Substituting the expressions for $\frac{dy}{dx}$ and $\frac{dx}{dt}$:

   $\frac{dy}{dt} = \frac{2x}{(x^2 + 1)^2} \cdot \frac{3}{2\sqrt{3t + 1}}$ Simplify: $\frac{dy}{dt} = \frac{3x}{(x^2 + 1)^2 \cdot \sqrt{3t + 1}}$

4. Substitute $x = \sqrt{3t + 1}$ into the result:

   Since $x = \sqrt{3t + 1}$, substitute this into the expression:

   $\frac{dy}{dt} = \frac{3\sqrt{3t + 1}}{((\sqrt{3t + 1})^2 + 1)^2 \cdot \sqrt{3t + 1}}$ Simplify further:

   $\frac{dy}{dt} = \frac{3}{(3t + 1 + 1)^2} = \frac{3}{(3t + 2)^2}$

Thus, the value of $\frac{dy}{dt}$ is:

$\frac{dy}{dt} = \frac{3}{(3t + 2)^2}$

Let me know if you would like further clarification or a more detailed explanation.

Here are some related questions you might find helpful:

1. How does the quotient rule work in differentiation?
2. What is the chain rule, and when is it used?
3. Can we find higher-order derivatives of the given function $y$?
4. How does implicit differentiation differ from the chain rule?
5. Can we apply a different method, like substitution, to solve this?

Tip: When differentiating composite functions, always identify the outer and inner functions clearly to apply the chain rule correctly.