I want to determine $f(x) = x+\sin x$ is homeomorphic or not on $\mathbb{R}$?
A bijective continuous function is homeomorphic if its inverse is also continuous. I know that $f$ is bijective. Also $f$ is continuous being the sum of two continuous functions.
How to look for the continuity of $f^{-1}$.
On
Once you know that $f$ is strictly increasing, just use this.
If you want to know some regularity of $f^{-1}$, continue reading.
Let $N=\{(2k+1)\pi:k\in\mathbb Z\}$.
In $\mathbb R\setminus N$ you have $f'\neq 0$, so $f$ is actually a diffeomorphism. Now we have to investigate what happens near $N$. By translation, it is sufficient to examine what happens around $\pi$. By Taylor expansion, we have that for $\epsilon>0$ small enough $$ |f(x)-f(\pi)| = |x+\sin(x)-\pi| \geq \left|\frac{(x-\pi)^3}{12}\right| \qquad \forall x\in(\pi-\epsilon,\pi+\epsilon) $$ so the inverse $f^{-1}$ is $\tfrac13$-Holder near $\pi$.
The function $f$ is continuous and strictly increasing, because its derivative is $\ge0$ and is positive on the intervals $(\pi+2k\pi,\pi+2(k+1)\pi)$.
This its inverse function exists and is strictly increasing as well and defined over $\mathbb{R}$ because $f$ is neither upper nor lower bounded. In particular $f^{-1}$ has left and right limit at every point, because $$ \lim_{x\to c^-}f^{-1}(x)=\sup\{f(x):x<c\} \qquad \lim_{x\to c^+}f^{-1}(x)=\inf\{f(x):x>c\} $$ Suppose that at some point $c$ the two limits are different, say $$ \lim_{x\to c^-}f^{-1}(x)=\sup\{f(x):x<c\}=a \qquad \lim_{x\to c^+}f^{-1}(x)=\inf\{f(x):x>c\}=b $$ with $a<b$. Then $(a+b)/2$ doesn't belong to the codomain of $f^{-1}$, which is the domain of $f$. Contradiction.