$\frac{dy}{dx} \;$ of $\; \ln\!\left(y\right)=e^{3y}\sin\!\left(x\right)$

Question

Find $dy/dx \;\;$ if $\;\; \ln\!\left(y\right)=e^{3y}\sin\!\left(x\right)$

So I did:

$\frac{dy}{dx} \frac{1}{y} = 3e^{3y} \frac{dy}{dx} \cdot \cos(x) $

moving $\frac{dy}{dx}$ to the right hand side:

$\frac{dy}{dx} (\frac{1}{y} - 3e^{3y}) = \cos(x)$

$\frac{dy}{dx} = \frac{\cos(x)}{\frac{1}{y} - 3e^{3y}}$

But the right answer seems to be:

$\frac{dy}{dx} = \frac{e^{3y} y \cos(x)}{1-3 e^{3 y} y sin(x)} $

Why is it so?

Answer 1

The derivative of product is not the product of derivatives.

$\frac{dy}{dx} \frac{1}{y} = 3e^{3y} \frac{dy}{dx} \cdot \sin(x) + e^{3y} \cdot\cos(x) $