\(y=vx => y/x=v\)

\(dy/dx= v+x (dv)/dx\)

\(v+x (dv)/dx= (2(vx)^2+x^2e^(-(v)^2))/(2x(vx))\)

\(v+x (dv)/dx= (2v^2x^2+x^2e^(-(v)^2))/(2vx^2)\)

\(v+x (dv)/dx= (x^2(2v^2+e^(-(v)^2)))/(2vx^2)\) (take \(x^2\) as common factor from numerator)

\(v+x (dv)/dx=(2v^2+e^(-(v)^2))/(2v)\) (divide \(x^2/x^2=1\))

Substract v from both sides and further simplify it

\(v+x (dv)/dx-v= (2v^2+e^(-(v)^2))/(2v)-v\)

\(x(dv)/dx= (2v^2e^(-(v)^2)-2v^2)/(2v)\)

\(x (dv)/dx= e^(-(v)^2)/(2x)\)

\((2v)/e^(-(v)^2) dv= dx/x\)

Integrate both sides

\(\int 2ve^{v^2}dv= \int dx/x+c\)

Substitute \(v^2=t\)

\(=> 2vdv=dt\)

Hence, \(\int e^t dt= \int dx/x+c\)

\(e^t= \ln x+c\)

\(t= \ln(\ln x+c)\) (use property \(e^x=y => x=\ln y\))

Substitute back \(t=v^2\)

\(v^2=\ln(\ln x+c)\)

Substitute back \(v=y/x\)

\((y/x)^2= \ln(\ln x+c)\)

\(y^2/x= \ln(\ln x+c)\)

\(y^2= x^2 \ln(\ln x+c)\)

\(y= +- \sqrt{x^2 \ln(\ln x+c)}\)

\(y= \pm x\sqrt{\ln (\ln x+C)}\)