$$ \frac{y}{x}\mathrm{d}x+(y^3+\ln{x})\mathrm{d}y = 0 $$
trying to apply the bernoulli method, dividing the whole expression into $y^3$ and
$$ \frac{1}{y^2x}\mathrm{d}x+(1+\frac{\ln{x}}{y^3})\mathrm{d}y=0 $$
to replace $ z = \frac{1}{y^2}$ and at this stage, stupor. can't separate parts from dx and dy
$$ \underbrace{\frac{y}{x}}_{M}\mathrm{d}x+\underbrace{(y^3+\ln{x})}_{N}\mathrm{d}y = 0 $$ The differential equation is exact means that $$ \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} $$ Indeed $$ \frac{\partial M}{\partial y} = \frac{\partial }{\partial y}\left(\frac{y}{x}\right) = \frac{1}{x} = \frac{\partial }{\partial x}(y^3+\ln{x}) = \frac{\partial N}{\partial x} $$ It follows that there is some function $u(x,y) = const$ such that $$ \frac{\partial u}{\partial x} = M, \qquad \frac{\partial u}{\partial y} = N $$ which also imlpies that the total differential of $u$ is the given differential equation \begin{align} du &= \frac{\partial u}{\partial x}dx + \frac{\partial u}{\partial y}dy \\ &= Mdx +Ndy \\ \end{align} The solution of it is the function $u(x,y) = const$.
