[tex]t\dfrac{\mathrm dy}{\mathrm dt}-y^2\ln t+y=0[/tex]
Divide both sides by [tex]y(t)^2[/tex]:
[tex]ty^{-2}\dfrac{\mathrm dy}{\mathrm dt}-\ln t+y^{-1}=0[/tex]
Substitute [tex]v(t)=y(t)^{-1}[/tex], so that [tex]\dfrac{\mathrm dv}{\mathrm dt}=-y(t)^{-2}\dfrac{\mathrm dy}{\mathrm dt}[/tex].
[tex]-t\dfrac{\mathrm dv}{\mathrm dt}-\ln t+v=0[/tex]
[tex]t\dfrac{\mathrm dv}{\mathrm dt}-v=\ln t[/tex]
Divide both sides by [tex]t^2[/tex]:
[tex]\dfrac1t\dfrac{\mathrm dv}{\mathrm dt}-\dfrac1{t^2}v=\dfrac{\ln t}{t^2}[/tex]
The left side can be condensed as the derivative of a product:
[tex]\dfrac{\mathrm d}{\mathrm dt}\left[\dfrac1tv\right]=\dfrac{\ln t}{t^2}[/tex]
Integrate both sides. The integral on the right side can be done by parts.
[tex]\displaystyle\int\frac{\ln t}{t^2}\,\mathrm dt=-\frac{\ln t}t+\int\frac{\mathrm dt}{t^2}=-\frac{\ln t}t-\frac1t+C[/tex]
[tex]\dfrac1tv=-\dfrac{\ln t}t-\dfrac1t+C[/tex]
[tex]v=-\ln t-1+Ct[/tex]
Now solve for [tex]y(t)[/tex].
[tex]y^{-1}=-\ln t-1+Ct[/tex]
[tex]\boxed{y(t)=\dfrac1{Ct-\ln t-1}}[/tex]
