<s>[INST] Determining $\sin(2x)$ 
Given that
$$\sin (y-x)\cos(x+y)=\dfrac 1 2$$
$$\sin (x+y)\cos (x-y) = \dfrac 1 3 $$
Determine $\sin (2x)$.

As stated in my perspective, the question does not make any sense. We know that the double angle identity for $\sin(2x)$ is given by 
$$\sin(2x) = 2\sin\cos$$
Let us try simpiflying the second equation
$$\sin(x+y)-\cos(x-y)=\sin x \cos y+\cos x \sin y-\cos x \cos y-\sin x\sin y=
\cos x(\sin y-\cos y)+\sin x(\cos y-\sin y)=\color{blue}{(\cos x-\sin x)(\sin y-\cos y})$$
However, there seems to be nothing useful. 
 [/INST] You were on the right track but the equations contain the product, not the difference.
\begin{align}
\frac12 &= \sin (y-x)\cos(x+y)\\
&=(\sin y\cos x - \sin x \cos y)(\cos x\cos y - \sin x\sin y) \\
&= \sin y\cos y - \cos x \sin x
\end{align}
\begin{align}
\frac13 &= \sin (x+y)\cos(x-y)\\
&=(\sin x\cos y + \sin y \cos x)(\cos x\cos y + \sin x\sin y) \\
&= \sin y\cos y + \cos x \sin x
\end{align}
So subtracting them gives
$$\sin 2x = 2\cos x\sin x = ( \sin y\cos y + \cos x \sin x) - ( \sin y\cos y - \cos x \sin x) = \frac13 - \frac12 = -\frac16$$
</s>