If [tex]t[/tex] is just a constant, then you have a fairly simple (separable) ODE.
[tex]t^2\dfrac{\mathrm dy}{\mathrm dx}+y^2=ty\implies \dfrac{\mathrm dy}{ty-y^2}=\dfrac{\mathrm dx}{t^2}[/tex]
and so on.
So, I'll assume you meant to write [tex]\dfrac{\mathrm dy}{\mathrm dt}[/tex]...
[tex]t^2\dfrac{\mathrm dy}{\mathrm dt}+y^2=ty[/tex]
This is a standard Bernoulli equation, which means a substitution of [tex]y=z^{1-2}=z^{-1}[/tex] will suffice to transform this ODE in [tex]y[/tex] into a linear ODE in [tex]z[/tex]. You have
[tex]\dfrac{\mathrm dy}{\mathrm dt}=\dfrac{\mathrm dy}{\mathrm dz}\times\dfrac{\mathrm dz}{\mathrm dt}[/tex]
[tex]\dfrac{\mathrm dy}{\mathrm dt}=-\dfrac1{z^2}\dfrac{\mathrm dz}{\mathrm dt}[/tex]
which changes the ODE to
[tex]-\dfrac{t^2}{z^2}\dfrac{\mathrm dz}{\mathrm dt}+\dfrac1{z^2}=\dfrac tz[/tex]
[tex]\dfrac{\mathrm dz}{\mathrm dt}+\dfrac1tz=\dfrac1{t^2}[/tex]
An integrating factor would be
[tex]\mu(t)=\exp\left(\displaystyle\int\frac{\mathrm dt}t\right)=t[/tex]
Multiplying both sides by the IF gives
[tex]t\dfrac{\mathrm dz}{\mathrm dt}+z=\dfrac1t[/tex]
[tex]\dfrac{\mathrm d}{\mathrm dt}[tz]=\dfrac1t[/tex]
Integrate both sides wrt [tex]t[/tex] to get
[tex]tz=\displaystyle\int\dfrac{\mathrm dt}t=\ln|t|[/tex]
[tex]z=\dfrac{\ln|t|+C}t[/tex]
Now back substitute. Since [tex]y=\dfrac1z[/tex], you get [tex]z=\dfrac1y[/tex] and the solution is
[tex]y=\dfrac t{\ln|t|+C}[/tex]
