Plugging in the first equation into the second, we get that $\sin ( \sin x ) = x$. Clear, $-1\leq x \leq 1$. We do casework on if $x$ is positive, negative or $0$. Case: $x$ is positive By graphing, it's clear that $\sin x > x$ for positive $x$. Also note that $\sin x$ is positive itself as $x$ is in the range $(0,1]$. So, $\sin (\sin x) > \sin x > x$ so there's no solutions for this case. Case: $x$ is negative By graphing, it's clear that $\sin x < x$ for negative $x$. Also note that $\sin x$ is negative itself as $x$ is in the range $[-1,0)$. So, $\sin (\sin x) < \sin x < x$. So, there's no solutions for this case. Case: $x=0$ Plugging in, $\sin(\sin 0)=\sin (0)= 0$, giving us a solution. The only solution is $\boxed{x=0,y=0}$