To prove the given equation, we will utilize the product-to-sum identities for cosine and sine. These identities are given as:

$\cos(a)\cos(b) = \dfrac{1}{2}[\cos(a-b) + \cos(a+b)]$$\sin(a)\sin(b) = \dfrac{1}{2}[\cos(a-b) - \cos(a+b)]$

Given Equation: $\cos(3x)\cos(2x) - \sin(x)\sin(6x) = \cos(7x)$

Expand the terms using the product-to-sum identities.

$\cos(3x)\cos(2x) = \dfrac{1}{2}[\cos(3x-2x) + \cos(3x+2x)]$
= $\dfrac{1}{2}[\cos(x) + \cos(5x)]$

$\sin(x)\sin(6x) = \dfrac{1}{2}[\cos(x-6x) - \cos(x+6x)]$
= $\dfrac{1}{2}[\cos(-5x) - \cos(7x)]$
= $-\dfrac{1}{2}[\cos(5x) + \cos(7x)]$

Therefore, $\cos(3x)\cos(2x) - \sin(x)\sin(6x) = \dfrac{1}{2}[\cos(x) + \cos(5x)] + \dfrac{1}{2}[\cos(5x) + \cos(7x)]$
= $\dfrac{1}{2}[\cos(x) + \cos(5x) + \cos(5x) + \cos(7x)]$
= $\dfrac{1}{2}[2\cos(5x) + \cos(x) + \cos(7x)]$
= $\cos(5x) + \dfrac{1}{2}[\cos(x) + \cos(7x)]$
= $\cos(5x) + \cos(6x)$

Hence, $\cos(3x)\cos(2x) - \sin(x)\sin(6x) = \cos(7x)$ has been proven correct.