The function is y = (2x) + 2(3x) + 2(x).

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

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

This can be written as $(\sin(3x))^2$.
Let $u = \sin(3x)$. Then $\frac{d}{dx}(u^2) = 2u \frac{du}{dx}$.
Now, find $\frac{du}{dx}$ for $u = \sin(3x)$.
Let $v = 3x$. Then $\frac{dv}{dx} = 3$.
The derivative of $\sin(v)$ is $\cos(v) \cdot \frac{dv}{dx}$.
So, $\frac{du}{dx} = \cos(3x) \cdot 3 = 3\cos(3x)$.
Now substitute back into the derivative of $u^2$:

$$\frac{d}{dx}(\sin^2(3x)) = 2\sin(3x) \cdot (3\cos(3x)) = 6\sin(3x)\cos(3x)$$

Using the identity $\sin(2\theta) = 2\sin\theta\cos\theta$, we can simplify this:

$$6\sin(3x)\cos(3x) = 3 \cdot (2\sin(3x)\cos(3x)) = 3\sin(2 \cdot 3x) = 3\sin(6x)$$

This can be written as $(\cos(x))^2$.
Let $w = \cos(x)$. Then $\frac{d}{dx}(w^2) = 2w \frac{dw}{dx}$.
Now, find $\frac{dw}{dx}$ for $w = \cos(x)$.

$$\frac{dw}{dx} = -\sin(x)$$

Now substitute back into the derivative of $w^2$:

$$\frac{d}{dx}(\cos^2(x)) = 2\cos(x) \cdot (-\sin(x)) = -2\sin(x)\cos(x)$$

Using the identity $\sin(2\theta) = 2\sin\theta\cos\theta$, we can simplify this:

$$-2\sin(x)\cos(x) = -\sin(2x)$$

$$\frac{dy}{dx} = \frac{d}{dx}(\tan(2x)) + \frac{d}{dx}(\sin^2(3x)) + \frac{d}{dx}(\cos^2(x))$$ $$\frac{dy}{dx} = 2\sec^2(2x) + 3\sin(6x) - \sin(2x)$$