1. **State the problem:** We need to solve the differential equation $$\frac{dy}{dx} = e^{x} \csc 2y \csc y$$ with the initial condition $$y(0) = \frac{\pi}{6}$$. 2. **Recall the formula and rules:** This is a separable differential equation. We can write it as: $$\frac{dy}{dx} = f(x)g(y)$$ which allows us to separate variables: $$\frac{dy}{g(y)} = f(x) dx$$ 3. **Rewrite the equation:** $$\frac{dy}{dx} = e^{x} \csc 2y \csc y$$ Separate variables: $$\frac{dy}{\csc 2y \csc y} = e^{x} dx$$ 4. **Simplify the left side:** Recall that $$\csc \theta = \frac{1}{\sin \theta}$$, so: $$\frac{1}{\csc 2y \csc y} = \sin 2y \sin y$$ Thus: $$\sin 2y \sin y \, dy = e^{x} dx$$ 5. **Use the double-angle identity:** $$\sin 2y = 2 \sin y \cos y$$ So: $$\sin 2y \sin y = 2 \sin y \cos y \sin y = 2 \sin^{2} y \cos y$$ 6. **Rewrite the separated equation:** $$2 \sin^{2} y \cos y \, dy = e^{x} dx$$ 7. **Integrate both sides:** $$\int 2 \sin^{2} y \cos y \, dy = \int e^{x} dx$$ 8. **Use substitution for the left integral:** Let $$u = \sin y$$, then $$du = \cos y \, dy$$. The integral becomes: $$\int 2 u^{2} du = 2 \int u^{2} du = 2 \cdot \frac{u^{3}}{3} + C = \frac{2}{3} \sin^{3} y + C$$ 9. **Integrate the right side:** $$\int e^{x} dx = e^{x} + C$$ 10. **Combine results:** $$\frac{2}{3} \sin^{3} y = e^{x} + C$$ 11. **Apply initial condition $$y(0) = \frac{\pi}{6}$$:** Calculate $$\sin \frac{\pi}{6} = \frac{1}{2}$$, so: $$\frac{2}{3} \left( \frac{1}{2} \right)^{3} = e^{0} + C \Rightarrow \frac{2}{3} \cdot \frac{1}{8} = 1 + C \Rightarrow \frac{1}{12} = 1 + C$$ 12. **Solve for $$C$$:** $$C = \frac{1}{12} - 1 = -\frac{11}{12}$$ 13. **Write the implicit solution:** $$\frac{2}{3} \sin^{3} y = e^{x} - \frac{11}{12}$$ 14. **Final answer:** $$\boxed{\frac{2}{3} \sin^{3} y = e^{x} - \frac{11}{12}}$$ This implicit solution relates $$y$$ and $$x$$ satisfying the differential equation and initial condition.