The question is to prove that $$\frac{d}{dx}[xJ_n(x)J_{n+1}(x)]=x[J_n^{2}(x)-J_{n+1}^{2}(x)]$$
I have attempted using the product rule and the definition of Bessel but I just end up with a mess and no end in sight. Any help would be greatly appreciated.
I think you didn't try very hard using the chain rule.
Our answer is supposed to be in terms of $J_n(x)$ and $J_{n+1}(x)$, so write everything in terms of them (use the recurrence to write $J_{n+2}$ in terms of $J_{n+1}$ and $J_n$).
$$ J_n'(x) = \frac{n}{x} J_n(x) - J_{n+1}(x) \tag{1} $$
$$ J_{n+1}'(x) = J_n(x) - \frac{n+1}{x} J_{n+1}(x) \tag{2} $$
Chain rule: $$ \frac{d}{dx}\big[xJ_n(x)J_{n+1}(x)\big] = J_n(x)J_{n+1}(x) +xJ_n'(x)J_{n+1}(x) +xJ_n(x)J_{n+1}'(x) $$ Plug in $(1)$ and $(2)$. Expand.