Since \(\sin y = \sin x\), we must have \[ y = x + 2 k \pi \] where \(k \in \ZZ \), or \[ y = (2k + 1) \pi - x \] where \(k \in \ZZ \).
For the first case, since \(x \in [-\pi , \pi ]\) and \(y \in [-\pi , \pi ]\), we must have simply \(x = y\).
For the second case, within this range, we can have \(y = \pi - x\), and \(y = - \pi - x\).
Hence, the sketch looks as follows.
Differentiating with respect to \(x\), we have \[ \cos y \DiffFrac {y}{x} = \frac {1}{2} \cos x. \]
Since \(\sin y = \frac {1}{2} \sin x\), \(\cos y = \pm \sqrt {1 - \sin ^2 y} = \pm \sqrt {1 - \frac {1}{4} \sin ^2 x}\). Since \(0 \leq y \leq \frac {1}{2} \pi \), \(\cos y > 0\), and hence \(\cos y = \frac {1}{2} \sqrt {4 - \sin ^2 x}\). Hence, \[ \DiffFrac {y}{x} = \frac {\frac {1}{2} \cos x}{\frac {1}{2} \sqrt {4 - \sin ^2 x}} = \frac {\cos x}{\sqrt {4 - \sin ^2 x}}. \]
Differentiating this again gives us \begin {align*} \NdiffFrac {2}{y}{x} & = \frac {(- \sin x) \sqrt {4 - \sin ^2 x} - \frac {1}{2} \cdot (-2 \sin x) \cdot \cos x \cdot \frac {1}{\sqrt {4 - \sin ^2 x}} \cdot \cos x}{4 - \sin ^2 x} \\ & = \frac {(- \sin x) (4 - \sin ^2 x) + \sin x \cos ^2 x}{(4 - \sin ^2 x)^{\frac {3}{2}}} \\ & = \frac {-4 \sin x + \sin ^3 x + \sin x (1 - \sin ^2 x)}{(4 - \sin ^2 x)^{\frac {3}{2}}} \\ & = -\frac {3 \sin x}{(4 - \sin ^2 x)^{\frac {3}{2}}}, \end {align*}
as desired.
Within this range of \(x\) and \(y\), we have \[ y = \arcsin \left (\frac {1}{2} \sin x\right ), \] and hence this is a function, and each \(x\) corresponds to a unique \(y\).
At \(x = 0\), \[ y = 0, y' = \frac {\cos 0}{\sqrt {4 - \sin ^2 0}} = \frac {1}{2}, \] and at \(x = \frac {\pi }{2}\), \[ y = \frac {\pi }{6}, y' = -\frac {\cos \frac {\pi }{2}}{\sqrt {4 - \sin ^2 \frac {\pi }{2}}} = 0. \]
Since \(y'' = - \frac {3 \sin x}{(4 - \sin ^2 x)^{\frac {3}{2}}} < 0\) for \(x \in \left [0, \frac {\pi }{2}\right ]\), this function is concave, and hence the graph looks as follows.
Hence, for \((x, y) \in [-\pi , \pi ]^2\), the graph looks as follows, by symmetry.
The graph is as follows.
