Answer
$$\frac{dy}{dx}=\frac{ye^y\cos x}{1-ye^y\sin x}$$
Work Step by Step
$$\ln y=e^y\sin x$$
We have $$(\ln x)'=\frac{1}{x}$$ $$(e^{x})'=e^x$$
Carrying out implicit differentiation on both sides, we have $$\frac{1}{y}\frac{dy}{dx}=e^y\frac{dy}{dx}\sin x+e^y\cos x$$
$$\frac{1}{y}\frac{dy}{dx}-e^y\sin x\frac{dy}{dx}=e^y\cos x$$
$$\frac{dy}{dx}\Big(\frac{1}{y}-e^y\sin x\Big)=e^y\cos x$$
$$\frac{dy}{dx}=\frac{e^y\cos x}{\frac{1}{y}-e^y\sin x}=\frac{e^y\cos x}{\frac{1-ye^y\sin x}{y}}$$
$$\frac{dy}{dx}=\frac{ye^y\cos x}{1-ye^y\sin x}$$
