Given:
$\frac{d^2y}{dx^2} \cdot \left(\tan(x) - 3\cos(x)\right) \cdot \frac{dy}{dx} + 2y \cos^2(x) = \cos^4(x)$

To solve this differential equation, we need to follow these steps:

Step 1: Simplify the equation
We can simplify the equation by expanding $\cos^4(x)$ using the binomial theorem:
$\cos^4(x) = (\cos^2(x))^2 = (\cos^2(x)) \cdot (\cos^2(x))$

Now the equation becomes:
$\frac{d^2y}{dx^2} \cdot \left(\tan(x) - 3\cos(x)\right) \cdot \frac{dy}{dx} + 2y \cos^2(x) = (\cos^2(x)) \cdot (\cos^2(x))$

Step 2: Substitute variables
Let's substitute $u = \cos(x)$, then $du = -\sin(x) dx$.
Now the equation becomes:
$\frac{d^2y}{dx^2} \cdot \left(\frac{\tan(x)}{-3u} - 3\right) \cdot \frac{dy}{dx} + 2y u^2 = u^2 \cdot u^2$

Step 3: Differentiate
Differentiating the equation with respect to $x$ will give us:
$\frac{d}{dx}\left(\frac{d^2y}{dx^2}\right) \cdot \left(\frac{\tan(x)}{-3u} - 3\right) \cdot \frac{dy}{dx} + \frac{d^2y}{dx^2} \cdot \left(\frac{d}{dx}\left(\frac{\tan(x)}{-3u} - 3\right)\right) \cdot \frac{dy}{dx} + 2y \cdot \frac{d}{dx}(u^2) = \frac{d}{dx}(u^2) \cdot u^2$

Simplifying the derivatives using the chain rule, we get:
$\frac{d^3y}{dx^3} \cdot \left(\frac{\tan(x)}{-3u} - 3\right) \cdot \frac{dy}{dx} - \frac{d^2y}{dx^2} \cdot \frac{\sec^2(x)}{-3u} \cdot \frac{dy}{dx} + 2y \cdot 2u \cdot \frac{du}{dx} = 2u \cdot u^2$

Step 4: Substitute variables and simplify
Substituting $du = -\sin(x) dx$ and $u = \cos(x)$, we get:
$\frac{d^3y}{dx^3} \cdot \left(\frac{\tan(x)}{-3u} - 3\right) \cdot \frac{dy}{dx} - \frac{d^2y}{dx^2} \cdot \frac{\sec^2(x)}{-3u} \cdot \frac{dy}{dx} +