Answer
(a)$$(f \circ g)(x)=\frac { \sin 2x}{1+ \sin 2x}$$ $$D_{f \circ g}=\mathbb R - \{x \mid x=\frac{3 \pi}{4}+n \pi , n \in \mathbb Z \}$$
(b)$$(g \circ f)(x)=\sin \left ( \frac{2x}{1+x} \right )$$ $$D_{g \circ f}=\mathbb R - \{ -1 \}$$
(c)$$(f \circ f)(x)=\frac{\frac{x}{1+x}}{1+ \frac{x}{1+x}}$$ $$D_{f \circ f}=\mathbb R - \{ -1, - \frac{1}{2} \}$$
(d)$$(g \circ g)(x)=\sin (2 \sin 2x )$$ $$D_{g \circ g}=\mathbb R$$
Work Step by Step
Domain of real rational function is the set of real number, $\mathbb R$, minus the points vanishing the denominator.
Domain of sine function is the set of real number, $\mathbb R$.
(a)$$(f \circ g)(x)= f(g(x))=\frac { \sin 2x}{1+ \sin 2x}$$ $$D_{f \circ g}= \mathbb R - \{ \text{roots of } (1+ \sin 2x =0) \} = \\ \mathbb R - \{ x \mid 2x= \frac{3 \pi}{2}+2n \pi, n \in \mathbb Z \}= \\ \mathbb R - \{x \mid x=\frac{3 \pi}{4}+n \pi , n \in \mathbb Z \}$$
(b)$$(g \circ f)(x)=g(f(x))= \sin \left ( \frac{2x}{1+x} \right )$$ $$D_{g \circ f}= \mathbb R - \{ \text{roots of } (1+x=0) \} = \\ \mathbb R - \{ -1 \}$$
(c)$$(f \circ f)(x)=f(f(x))=\frac{\frac{x}{1+x}}{1+ \frac{x}{1+x}}$$ $$D_{f \circ f}= \mathbb R - \left ( \{ \text{roots of } (1+x=0) \} \cup \left \{ \text{roots of } \left ( 1+ \frac{x}{1+x} =0 \right ) \right \} \right )=\mathbb R - \left ( \{ \text{roots of } (1+x=0) \} \cup \{ \text{roots of } (2x+1=0) \} \right ) \\= \mathbb R - \{ -1, - \frac{1}{2} \}$$For $x \in D_{f \circ f}$, we can write this composition function as $$(f \circ f)(x)= \frac{ \frac{x}{1+x}}{ \frac{2x+1}{1+x}}= \frac{x}{2x+1}.$$
(d)$$(g \circ g)(x)=g(g(x))= \sin (2 \sin 2x )$$ $$D_{g \circ g}= \mathbb R$$
