We start with the Left Hand Side (L.H.S.) of the given expression:

$$\text{L.H.S.} = \frac{\sin x - \sin 3x}{\sin^2 x - \cos^2 x}$$

To simplify the numerator, we use the identity for $\sin 3x$ provided in the context:

$$\sin 3x = 3 \sin x - 4 \sin^3 x$$

Substituting this into the numerator, we get:

$$\sin x - (3 \sin x - 4 \sin^3 x) = \sin x - 3 \sin x + 4 \sin^3 x$$ $$= -2 \sin x + 4 \sin^3 x = 2 \sin x (2 \sin^2 x - 1)$$

For the denominator, we rearrange terms to match the standard identity for $\cos 2x$:

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

From the context, we know $\cos 2x = \cos^2 x - \sin^2 x$. Thus, the denominator is $-\cos 2x$.

We also use the alternative form for $\cos 2x$ from the context: $\cos 2x = 1 - 2 \sin^2 x$. Substituting this into the denominator expression:

$$-\cos 2x = -(1 - 2 \sin^2 x) = 2 \sin^2 x - 1$$

Now, dividing the simplified numerator by the simplified denominator:

$$\text{L.H.S.} = \frac{2 \sin x (2 \sin^2 x - 1)}{2 \sin^2 x - 1}$$

Cancelling the common term $(2 \sin^2 x - 1)$, we get:

$$\text{L.H.S.} = 2 \sin x = \text{R.H.S.}$$

Hence proved.