Solve the differential equation: dy/dx = (3x2 * sin(x3)) / cos(y)

Rearrange the differential equation to group terms involving $y$ with $dy$ and terms involving $x$ with $dx$.

$$\cos(y) \frac{dy}{dx} = 3x^2 \sin(x^3)$$ $$\cos(y) dy = 3x^2 \sin(x^3) dx$$

For the left side:

$$\int \cos(y) dy = \sin(y) + C_1$$

For the right side, use a substitution. Let $u = x^3$. Then $du = 3x^2 dx$.

$$\int \sin(u) du = -\cos(u) + C_2$$

Substitute back $u = x^3$:

$$\int 3x^2 \sin(x^3) dx = -\cos(x^3) + C_2$$

$$\sin(y) + C_1 = -\cos(x^3) + C_2$$

Let $C = C_2 - C_1$ be the arbitrary constant.

$$\sin(y) = -\cos(x^3) + C$$

To solve for $y$, take the inverse sine of both sides:

$$y = \arcsin(-\cos(x^3) + C)$$