Differentiate the first term, x.

The derivative of $\sin x$ with respect to $x$ is $\cos x$.

$$\frac{d}{dx}(\sin x) = \cos x$$

The product rule states that if $u(x) = f(x)g(x)$, then $\frac{du}{dx} = f'(x)g(x) + f(x)g'(x)$.
Let $f(x) = x^3$ and $g(x) = \cos x$.
Then, $f'(x) = 3x^2$ and $g'(x) = -\sin x$.
Applying the product rule:

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

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

$$\frac{dy}{dx} = (1 - 3x^2) \cos x + x^3 \sin x$$

Comparing this accurate result with the given options:
a. $(1-3x)\cos x + x^2 \sin x$
b. $(1-3x)\cos x + x^3 \sin x$
c. $(1+3x)\cos x + x^3 \sin x$
d. $(1-3x)\cos x - x^3 \sin x$
e. $(1-2x)\cos x + x^3 \sin x$

The mathematically correct derivative is $(1 - 3x^2) \cos x + x^3 \sin x$. None of the provided options exactly match this result. Option b, $(1-3x)\cos x + x^3 \sin x$, is the closest, differing only in the power of $x$ in the coefficient of $\cos x$ (it has $3x$ instead of $3x^2$).

$$\boxed{(1 - 3x^2) \cos x + x^3 \sin x}$$