The problem asks for the equation of the tangent line to the curve $y = \tan^2(x)$ at the point $\left(\frac{\pi}{3}, 3 \right)$.

We need to:

1. Differentiate $y = \tan^2(x)$ to find the slope of the tangent line at $x = \frac{\pi}{3}$.
2. Use the point-slope form of a line to write the equation.

Step 1: Find the derivative of $y = \tan^2(x)$

Using the chain rule: $\frac{dy}{dx} = 2 \tan(x) \cdot \sec^2(x)$ This is the derivative of $y = \tan^2(x)$.

Step 2: Evaluate the derivative at $x = \frac{\pi}{3}$

First, calculate $\tan\left(\frac{\pi}{3}\right)$ and $\sec\left(\frac{\pi}{3}\right)$: $\tan\left(\frac{\pi}{3}\right) = \sqrt{3}, \quad \sec\left(\frac{\pi}{3}\right) = 2$ Now substitute into the derivative formula: $\frac{dy}{dx} \Big|_{x = \frac{\pi}{3}} = 2 \cdot \sqrt{3} \cdot (2^2) = 2 \cdot \sqrt{3} \cdot 4 = 8\sqrt{3}$ So, the slope $m = 8\sqrt{3}$.

Step 3: Find the equation of the tangent line

We use the point-slope form of a line: $y - y_1 = m(x - x_1)$ where $m = 8\sqrt{3}$, $x_1 = \frac{\pi}{3}$, and $y_1 = 3$. Substituting these values: $y - 3 = 8\sqrt{3} \left( x - \frac{\pi}{3} \right)$ Simplifying this into slope-intercept form $y = mx + b$, we get: $y = 8\sqrt{3}x - 8\sqrt{3}\frac{\pi}{3} + 3$ $y = 8\sqrt{3}x - \frac{8\pi\sqrt{3}}{3} + 3$ So, $m = 8\sqrt{3}$ and $b = 3 - \frac{8\pi\sqrt{3}}{3}$.

Let me know if you'd like further clarifications or additional steps.

Relative questions:

1. How does the chain rule apply in differentiating $\tan^2(x)$?
2. What is the point-slope form of a line, and why do we use it for tangent lines?
3. Why is the derivative of $\tan(x)$ related to $\sec^2(x)$?
4. How can we simplify terms involving $\pi$ in the final answer for $b$?
5. What would happen if we changed the function to $\cot(x)$ instead of $\tan(x)$?

Tip: When solving problems involving the tangent line, always start by differentiating to find the slope before applying the point-slope formula.