Answer
$$y= t e^{\left(1-\frac{1}{t}\right)}-1$$
Work Step by Step
Given
$$t^{2} \frac{d y}{d t}-t=1+y+t y, \quad y(1)=0$$
Rewrite the equation
\begin{align*}
t^{2} \frac{d y}{d t}&=1+t+y+t y\\
&=(1+t)(1+y)
\end{align*}
Separating the variables, we get
\begin{align*}
\frac{d y}{1+y}&=\frac{1+t}{t^{2} }dt\\
\int \frac{d y}{1+y}&= \int t^{-2}+ \frac{1 }{t }dt\\
\ln |1+y|&= \frac{-1}{t}+\ln t+C
\end{align*}
Since $y(1)=0$, then $C= 1$, and
\begin{align*}
\ln |1+y|&= \frac{-1}{t}+\ln t+1\\
\ln \frac{1+y}{t}&=1-\frac{1}{t}\\
\frac{1+y}{t}&=e^{1-\frac{1}{t} }\\
y&= t e^{\left(1-\frac{1}{t}\right)}-1
\end{align*}
