The Bernouli's equation

$\frac{d y}{d x}-y \tan x=\frac{\sin x \cos ^2 x}{y^2}$ can be transformed to

$\frac{d z}{d x}-3 z \tan x=3 \sin x \cos ^2 x$

We have,

$y^2 \frac{d y}{d x}-y^3 \tan x=\sin x \cos ^2 x$

Putting $y^3=z$ and $y^2 \frac{d y}{d x}=\frac{1}{3} \frac{d z}{d x}$, we get

$\frac{1}{3} \frac{d z}{d x}-z \tan x=\sin x \cos ^2 x$

$\Rightarrow \frac{d z}{d x}-3 z \tan x =3 \sin x \cos ^2 x$