$ \frac{dy}{dx}=xy+y^2e^{-\frac{x^2}{2}}\log x$

Dividing throughout by $y^2$,we get

$\frac{1}{y^2}\frac{dy}{dx}-\frac{1}{y}x=e^{-\frac{x^2}{2}\log x} $ $\dots\dots(1)$

put

$\frac{-1}{y}=v$

$ \frac{1}{y^2}\frac{dy}{dx}=\frac{dv}{dx}$

Eqn(1)becomes

$ \frac{dv}{dx}+xv=e^{-\frac{x^2}{2}\log x} $

This is a linear differential equation in $v $with $P=x$ and $Q=e^{-\frac{x^2}{2}}\log x$

$I.F=e^{\int xdx}=e^{\frac{x^2}{2}} $

The solution is

$ve^{\frac{x^2}{2}}=\int e^{\frac{x^2}{2}}e^{\frac{-x^2}{2}}\log xdx+c$

$ ve^{\frac{x^2}{2}}=\int \log xdx+c$

$e^{\frac{x^2}{2}}=\log x.x-\int x.\frac{1}{x}dx+c$

$ve^{\frac{x^2}{2}}=\log x.x-x+c$

$ -\frac{1}{y}e^{\frac{x^2}{2}}=x(\log x-1)+c$