Rewrite the function by splitting the fraction and using trigonometric identities.

$$y = \frac{\sin(x^2)}{\cos(x^2)} - \frac{3\tan(p^2)}{\cos(x^2)}$$ $$y = \tan(x^2) - 3\tan(p^2) \cdot \sec(x^2)$$

Let $u = x^2$. Then $\frac{du}{dx} = 2x$.
The derivative of $\tan(u)$ is $\sec^2(u) \cdot \frac{du}{dx}$.

$$\frac{d}{dx}(\tan(x^2)) = \sec^2(x^2) \cdot 2x = 2x\sec^2(x^2)$$

Since $p$ is a constant, $3\tan(p^2)$ is also a constant. Let $C = 3\tan(p^2)$.
We need to find $\frac{d}{dx}(-C \sec(x^2))$.
The derivative of $\sec(u)$ is $\sec(u)\tan(u) \cdot \frac{du}{dx}$.
Here, $u = x^2$, so $\frac{du}{dx} = 2x$.

$$\frac{d}{dx}(-C \sec(x^2)) = -C \cdot (\sec(x^2)\tan(x^2) \cdot 2x)$$

Substitute $C = 3\tan(p^2)$ back:

$$= -3\tan(p^2) \cdot 2x\sec(x^2)\tan(x^2)$$ $$= -6x\tan(p^2)\sec(x^2)\tan(x^2)$$

$$\frac{dy}{dx} = 2x\sec^2(x^2) - 6x\tan(p^2)\sec(x^2)\tan(x^2)$$

The final answer is $\boxed{\frac{dy}{dx} = 2x\sec^2(x^2) - 6x\tan(p^2)\sec(x^2)\tan(x^2)}$.