For
$$y' = f_0(x) +f_1(x)*y + f_2(x) *y^2 \hspace{5mm} \text{for} \, f_2(x) \not\equiv 0$$
then, suppose $q_{0}(x)$ is a solution, let $y(x) = q_{0}(x) + p(x)$ to obtain
\begin{align}
q_{0}' + p' &= f_{0} + f_{1} q_{0} + f_{1} p + f_{2} (q_{0}^{2} + 2 q_{0} p + p^{2}) \\
p' &= f_{2} p^{2} + (f_{1} + 2 q_{0} f_{2}) p + (- q_{0}^{'} + f_{2} q_{0}^{2} + f_{1} q_{0} + f_{0}) \\
p' &= f_{2} p^{2} + Q_{0} p,
\end{align}
where $Q_{0} = f_{1} + 2 q_{0} f_{2}$.
Now
\begin{align}
\frac{p'}{p^{2}} - \frac{Q_{0}}{p} &= f_{2} \\
- \frac{d}{dx} \left(\frac{1}{p}\right) - Q_{0} \, \frac{1}{p} &= f_{2}
\end{align}
Let $v(x) = 1/p(x)$ to obtain, with use of integrating factor,
\begin{align}
v' + Q_{0} v &= - f_{2} \\
\frac{d}{dx} \left[ v \, e^{\int^{x} Q_{0} dt} \right] &= - f_{2} \, e^{ \int^{x} Q_{0} dt} \\
v &= e^{- \int^{x} Q_{0} dt} \, \left[c - \int^{x} f_{2}(u) \, e^{\int^{u} Q_{0} dt} \, du \right]
\end{align}
Placing everything together provides
$$y(x) = q_{0}(x) + e^{\int^{x} (f_{1}(t) + 2 q_{0}(t) f_{2}(t)) dt} \, \left(c + \int^{x} f_{2}(u) \, e^{- \int^{u} (f_{1}(t) + 2 q_{0}(t) f_{2}(t)) dt} \, du \right)^{-1} $$