Non-linear but Trivial?

We can see that if $z=xy$ , the equation becomes $z'=z^2\ln{x}$ , which is separable; its solution is given by $-\frac{1}{xy}=-\frac{1}{z}=\int\frac{\mathrm{d}z}{z^2}=\int{\ln{x}\ \mathrm{d}x}=x(\ln{x}-1)+C\,.$

Substituting for $x=1$ , we find out that $-3=-1+C$ , so $C=-2$ . Substituting $x=e$ , we find out that $-\frac{1}{ey}=C=-2$ , so $y=\frac{1}{2e}$ .

The function comes out to be :

$\frac{1}{xy} + x(ln(x)-1) = 2$

Solution for this problem :

Rewriting the differential equation we get :

$\large x\frac{{y}^{'}}{{y}^{2}} + \frac{1}{y} = {x}^{2}(ln(x))$

Put $\frac{1}{y} = t$

Our differential equation becomes :

$\large -x{t}^{'}+t={x}^{2}ln(x)$

$\large \frac{dt}{dx}-\frac{t}{x}=-xln(x)$

Now this a linear differential equation, hence we find our integrating factor and that comes out to be :

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

Multiplying by integrating factor we get :

$\large \frac { 1 }{ x } \frac { dt }{ dx } -\frac { t }{ { x }^{ 2 } } =-ln(x)$

Integrating both sides we get :

$\large \frac { t }{ x } =x(1-ln(x))+C$

Putting back $t=\frac{1}{y}$ we get :

$\large \frac { 1 }{ xy } =x(1-ln(x))+C$

Put $x=1,y=\frac{1}{3}$ to get $C=2$

Hence our solution is :

$\large \frac { 1 }{ xy } +x(ln(x)-1)=2$