The first-order ODE $$\frac{\dd y}{\dd t}=\frac{-9 t^2+2 t + 3}{2y}$$ can be rewritten in differential form as $$2y\,\dd y=(-9 t^2+2 t + 3)\,\dd t$$ The left- and right-hand sides can be replaced by $\dd(y^2)$ and $\dd(-3 t^3+t^2+3 t)$, respectively. After all, $y^2$ and $-3 t^3+t^2+3 t$ are antiderivatives of $y$ and $-9 t^2+2 t + 3$, respectively. So we have an equality between two differentials: $$\dd(y^2)=\dd(-3 t^3+t^2+3 t)$$ The functions ‘behind the d's’ are thus equal to each other up to a constant: $$y^2=-3 t^3+t^2+3 t+c$$ for some constant $c$. The explicit solution is: $$y=\sqrt{-3 t^3+t^2+3 t+c}\quad \text{or} \quad y=-\sqrt{-3 t^3+t^2+3 t+c}$$

For those who find the final steps in the processing of differentials too formal or are not familar with differentials: you can also add to the left- and right-hand side of the $$2y\,\dd y=(-9 t^2+2 t + 3)\,\dd t$$ the integral symbol: $$\int 2y\,\dd y=\int (-9 t^2+2 t + 3)\,\dd t$$ Now you just continue with computing these integrals. So: $$y^2=-3 t^3+t^2+3 t+c$$ where we have combined the integration constants into a single constant $c$ and have placed it on the right-hand side of the equation.