Let $S_0=\{z\in\mathbb C:|z|=1,z\ne-1\}$ and $f(z)=\frac{\operatorname{Im}z}{1+\operatorname{Re}z}$. Prove that $f$ is a bijection between $S_0$ and $\mathbb R$. Find $f^{-1}$.
1. **Define the set \( S_0 \) and the function \( f \):**
   \[
   S_0 = \{ z \in \mathbb{C} : |z| = 1, z \neq -1 \}
   \]
   \[
   f(z) = \frac{\operatorname{Im} z}{1 + \operatorname{Re} z}
   \]

2. **Express \( z \) in polar form:**
   Since \( |z| = 1 \), we can write \( z \) as \( z = e^{i\theta} \) where \( \theta \in (-\pi, \pi) \setminus \{-\pi\} \).

3. **Compute \( \operatorname{Re} z \) and \( \operatorname{Im} z \):**
   \[
   \operatorname{Re} z = \cos \theta, \quad \operatorname{Im} z = \sin \theta
   \]

4. **Substitute into \( f(z) \):**
   \[
   f(z) = \frac{\sin \theta}{1 + \cos \theta}
   \]

5. **Simplify using trigonometric identities:**
   Using the identity \( 1 + \cos \theta = 2 \cos^2(\theta/2) \) and \( \sin \theta = 2 \sin(\theta/2) \cos(\theta/2) \):
   \[
   f(z) = \frac{2 \sin(\theta/2) \cos(\theta/2)}{2 \cos^2(\theta/2)} = \frac{\sin(\theta/2)}{\cos(\theta/2)} = \tan(\theta/2)
   \]

6. **Show that \( f \) is a bijection:**
   - **Injectivity:** Suppose \( f(z_1) = f(z_2) \). Then \( \tan(\theta_1/2) = \tan(\theta_2/2) \). Since \( \theta \in (-\pi, \pi) \), it follows that \( \theta_1 = \theta_2 \) and hence \( z_1 = z_2 \).
   - **Surjectivity:** For any \( y \in \mathbb{R} \), there exists \( \theta \in (-\pi, \pi) \) such that \( \tan(\theta/2) = y \). Thus, \( f(e^{i\theta}) = y \).

7. **Find the inverse function \( f^{-1} \):**
   Given \( y = \tan(\theta/2) \), we need to express \( e^{i\theta} \) in terms of \( y \).

8. **Use trigonometric identities to find \( \cos \theta \) and \( \sin \theta \):**
   \[
   \tan(\theta/2) = y \implies \theta/2 = \tan^{-1}(y)
   \]
   \[
   \cos(\theta/2) = \frac{1}{\sqrt{1+y^2}}, \quad \sin(\theta/2) = \frac{y}{\sqrt{1+y^2}}
   \]
   \[
   \cos \theta = 1 - 2 \sin^2(\theta/2) = \frac{1 - y^2}{1 + y^2}
   \]
   \[
   \sin \theta = 2 \sin(\theta/2) \cos(\theta/2) = \frac{2y}{1 + y^2}
   \]

9. **Express \( z \) in terms of \( y \):**
   \[
   f^{-1}(y) = \cos \theta + i \sin \theta = \frac{1 - y^2}{1 + y^2} + i \frac{2y}{1 + y^2}
   \]

Conclusion:
\[
\boxed{f^{-1}(y) = \frac{1 - y^2}{1 + y^2} + i \frac{2y}{1 + y^2}}
\]