To solve for $\frac{dy}{dx}$ and find $f(x)$, we start with the given equation:

$x = \ln(\sin y), \quad 0 < y < 2\pi$

Step 1: Differentiate both sides with respect to $x$

Using implicit differentiation, we differentiate both sides with respect to $x$:

$\frac{d}{dx} \left( x \right) = \frac{d}{dx} \left( \ln(\sin y) \right)$

$1 = \frac{1}{\sin y} \cdot \cos y \cdot \frac{dy}{dx}$

This simplifies to:

$1 = \cot y \cdot \frac{dy}{dx}$

$\frac{dy}{dx} = \frac{1}{\cot y} = \tan y$

Step 2: Express $\tan y$ in terms of $x$

From the original equation $x = \ln(\sin y)$, we can rewrite $\sin y$ as:

$\sin y = e^x$

Since $\cos^2 y + \sin^2 y = 1$, the cosine can be expressed as:

$\cos y = \sqrt{1 - \sin^2 y} = \sqrt{1 - e^{2x}}$

Thus, $\tan y = \frac{\sin y}{\cos y}$ becomes:

$\tan y = \frac{e^x}{\sqrt{1 - e^{2x}}}$

Step 3: Write $\frac{dy}{dx}$ in the required form

Substitute this result into $\frac{dy}{dx}$:

$\frac{dy}{dx} = \tan y = \frac{e^x}{\sqrt{1 - e^{2x}}}$

Comparing this with $\frac{dy}{dx} = \frac{e^x}{f(x)}$, we identify $f(x)$ as:

$f(x) = \sqrt{1 - e^{2x}}$

Final Answer:

$\frac{dy}{dx} = \frac{e^x}{\sqrt{1 - e^{2x}}}$

where $f(x) = \sqrt{1 - e^{2x}}$.

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Would you like further details on any step or a different explanation? Here are 5 related questions you might explore:

1. What happens to $\frac{dy}{dx}$ as $x$ approaches $-\infty$ or $\ln(1)$?
2. How can we verify that $f(x)$ is always real for the given domain of $x$?
3. Can this curve be parametrized explicitly as $y$ in terms of $x$?
4. What is the domain of $x$ based on the restriction $0 < y < 2\pi$?
5. How does the behavior of $\tan y$ affect the slope of the curve?

Tip: Always consider the domain of trigonometric and logarithmic functions carefully when transforming equations.