Let \(f(x) = \sin(\sin x)\).
Domain of definition, range and intersections with axes: The domain of definition is all real numbers. To determine the function's range we first note that the range of the internal \(\sin x\) is the interval \([-1{,}1]\). So the range of \(f\) is the image of the interval \([-1, 1]\) under the (outer) sine, which is the interval \([\sin -1, \sin 1]\) since the sine function is increasing and continuous on the interval \([-1{,}1]\). The intersection of the function's graph with the \(y\) axis is the point \((0, \sin(\sin(0)) = (0{,}0)\). For intersection points with the \(x\) axis we must solve the equation \(\sin(\sin x) = 0\). The outer sine is equal to \(0\) whenever the inner sine is equal to \(k \pi\) for \(k \in \mathbb Z\). Since the inner sine takes on values in the range \([-1, 1]\) the only possibility is \(\sin (x) = 0\), which happens when \(x = k \pi\).
Other imporatant characteristics: Before we begin to differentiate, we note that \(f\) is a \(2\pi\)-periodic function, so it suffices to investigate it on the interval \([0, 2\pi]\) and its behavior elsewhere follows from periodicity. Also \(f\) is odd, so it is symmetric about the origin.
Monotonicity:
We compute \(f'(x) = \cos(\sin x)\cdot \cos x\). Since \(\sin x \in [-1{,}1]\) we know that \(\cos(\sin x)\) is always positive. So the sign of \(\cos(\sin x)\cdot \cos x\) is determined by the sign of \(\cos x\). On the interval \([0, 2\pi]\), \(\cos x\) is positive on \([0, \pi/2) \cup ((3/2)\pi, 2\pi]\), negative on \((\pi/2, (3/2)\pi)\) and zero at the points \(\pi/2\) a \((3/2)\pi\). So we know that \(f\) is increasing on \([0, \pi/2)\) and \(((3/2)\pi, 2\pi]\) and decreasing on \((\pi/2, (3/2)\pi)\). And so since \(f\) is continuous, it has a local maximum at \(x = \pi/2\) and a local minimum at \(x = (3/2)\pi\).
Limits at boundary points: We can easily realize that neither \(\lim_{x\to \infty} f(x)\) nor \(\lim_{x \to -\infty}\) exist. This more or less follows from the fact that the function is periodic. To be more precise, if we let \(x_k = 2k\pi\) and \(y_k = \pi/2 + 2k\pi\), we obtain \(f(x_k) = 0\) and \(f(y_k) = \sin(1)\). So we have two sequences (\(x_k, y_k \to \infty\)), and if we apply \(f\) to each of them, we obtain different limts. So by Heine's condition \(\lim_{x\to \infty} f(x)\) cannot exist. The situation with the second limit is analogous.
Asymptotes: The function has no asymptotes, which follows from the fact that the limits \(\lim_{x \to \pm \infty} f(x)\) do not exist. If, for instance, there were an asymptote at \(+\infty\), then \(\lim_{x \to \infty} f(x)\) would have to be either \(\pm \infty\) (if the asymptote had a non-zero slope) or a finite value (if the asymptote had a slope of zero).
Now we can finally sketch the graph, even though we don't precisely know the inflection points:
