We begin by separating the variables and writing
\begin{equation*}
y^2\frac{dy}{dt} = t\text{.}
\end{equation*}
Integrating both sides of the equation with respect to the independent variable \(t\) shows that
\begin{equation*}
\int y^2\frac{dy}{dt}~dt = \int t~dt\text{.}
\end{equation*}
Next, we notice that the left-hand side allows us to change the variable of antidifferentiation from \(t\) to \(y\text{.}\) In particular, \(dy = \frac{dy}{dt}~dt\text{,}\) so we now have
\begin{equation*}
\int y^2 ~dy = \int t~dt\text{.}
\end{equation*}
This equation says that two families of antiderivatives are equal to each other. Therefore, when we find representative antiderivatives of both sides, we know they must differ by an arbitrary constant \(C\text{.}\) Antidifferentiating and including the integration constant \(C\) on the right, we find that
\begin{equation*}
\frac{y^3}{3} = \frac{t^2}{2} + C\text{.}
\end{equation*}
It is not necessary to include an arbitrary constant on both sides of the equation; we know that \(y^3/3\) and \(t^2/2\) are in the same family of antiderivatives and must therefore differ by a single constant.
Finally, we solve the last equation above for \(y\) as a function of \(t\text{,}\) which gives
\begin{equation*}
y(t) = \sqrt[3]{\frac 32 \, t^2 + 3C}\text{.}
\end{equation*}
Of course, the term \(3C\) on the right-hand side represents 3 times an unknown constant. It is, therefore, still an unknown constant, which we will rewrite as \(C\text{.}\) We thus conclude that the function
\begin{equation*}
y(t) = \sqrt[3]{\frac 32 \, t^2 + C}
\end{equation*}
is a solution to the original differential equation for any value of \(C\text{.}\)
Notice that because this solution depends on the arbitrary constant \(C\text{,}\) we have found an infinite family of solutions. This makes sense because we expect to find a unique solution that corresponds to any given initial value.
For example, if we want to solve the initial value problem
\begin{equation*}
\frac{dy}{dt} = \frac{t}{y^2}, \ y(0) = 2\text{,}
\end{equation*}
we know that the solution has the form \(y(t) = \sqrt[3]{\frac32\, t^2 + C}\) for some constant \(C\text{.}\) We therefore must find the appropriate value for \(C\) that gives the initial value \(y(0)=2\text{.}\) Hence,
\begin{equation*}
2 = y(0) = \sqrt[3]{\frac 32 \, 0^2 + C} = \sqrt[3]{C}\text{,}
\end{equation*}
which shows that \(C = 2^3 = 8\text{.}\) The solution to the initial value problem is then
\begin{equation*}
y(t) = \sqrt[3]{\frac32\, t^2+8}\text{.}
\end{equation*}