\[ f(x) y(x)^2+g(x) y(x)+y'(x)=0 \] ✓ Mathematica : cpu = 0.101978 (sec), leaf count = 54
\[\left \{\left \{y(x)\to \frac {\exp \left (\int _1^x-g(K[1])dK[1]\right )}{-\int _1^x-\exp \left (\int _1^{K[2]}-g(K[1])dK[1]\right ) f(K[2])dK[2]+c_1}\right \}\right \}\] ✓ Maple : cpu = 0.03 (sec), leaf count = 28
\[y \relax (x ) = \frac {{\mathrm e}^{\int -g \relax (x )d x}}{\int {\mathrm e}^{\int -g \relax (x )d x} f \relax (x )d x +c_{1}}\]
Hand solution
\begin {align} y^{2}f+gy+y^{\prime } & =0\nonumber \\ y^{\prime } & =-gy-y^{2}f\nonumber \\ & =P\relax (x) +Q\relax (x) y+R\relax (x) y^{2}\tag {1} \end {align}
This is Bernoulli first order non-linear ODE. \(P\relax (x) =0,Q\left ( x\right ) =-g,R\relax (x) =f\). First step is to divide by \(y^{2}\)\begin {equation} \frac {y^{\prime }}{y^{2}}=-g\frac {1}{y}-f\tag {2} \end {equation}
Let \(u=\frac {1}{y}\), then \(u^{\prime }=\frac {-y^{\prime }}{y^{2}}\) and (2) becomes\begin {align*} -u^{\prime } & =-gu-f\\ u^{\prime }-gu & =f \end {align*}
Integrating factor is \(e^{-\int gdx}\) hence\begin {align*} d\left (e^{-\int gdx}u\right ) & =fe^{-\int gdx}\\ e^{-\int gdx}u & =\int fe^{-\int gdx}dx+C\\ u & =e^{\int gdx}\left (\int fe^{-\int gdx}dx+C\right ) \end {align*}
Hence \begin {align*} y & =\frac {1}{e^{\int gdx}\left (\int fe^{-\int gdx}+C\right ) }\\ & =\frac {e^{-\int gdx}}{\int fe^{-\int gdx}dx+C} \end {align*}
Let \(\beta =e^{-\int gdx}\) then\[ y=\frac {\beta }{\int f\beta dx+C}\]
Verification