The solution of $\mathrm{x}\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{y}={\mathrm{y}}^2\log \mathrm{x}$ is

$$\begin{gathered} \mathrm{x}\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{y}={\mathrm{y}}^2\log \mathrm{x} \\ \Rightarrow {\mathrm{y}}^{-2}\frac{\mathrm{dy}}{\mathrm{dx}}+\frac{1}{\mathrm{x}}{\mathrm{y}}^{-1}=\frac{1}{\mathrm{x}}\log \mathrm{x} \end{gathered}$$

put y^–1 = t 

$\Rightarrow -y^{-2}\frac{dy}{dx}=\frac{dt}{dx}$

$\Rightarrow \frac{dt}{dx}-\frac{t}{x}=-\frac{\log x}{x}$

I.F.$=e^{\int \frac{-1}{x}dx}=\frac{1}{x}$

$\Rightarrow \frac{t}{x}=-\int \frac{1}{x}\times \frac{\log \left(x\right)}{x}+c$

$\Rightarrow$$\frac{1}{xy}=\frac{\log x}{x}+\frac{1}{x}+c$