\[ y(x) f'(x)-f(x) f'(x)+y'(x)=0 \] ✓ Mathematica : cpu = 0.023509 (sec), leaf count = 18
\[\left \{\left \{y(x)\to f(x)+c_1 e^{-f(x)}-1\right \}\right \}\] ✓ Maple : cpu = 0.022 (sec), leaf count = 15
\[y \relax (x ) = f \relax (x )-1+{\mathrm e}^{-f \relax (x )} c_{1}\]
Hand solution
\begin {equation} \frac {dy}{dx}+y\relax (x) \frac {df}{dx}=f\relax (x) \frac {df}{dx} \tag {1} \end {equation}
Integrating factor \(\mu =e^{\int \frac {df}{dx}dx}=e^{f}\). Therefore (1) becomes\[ \frac {d}{dx}\left (e^{f}y\relax (x) \right ) =e^{f}f\relax (x) \frac {df}{dx}\] Integrating\begin {align*} e^{f}y\relax (x) & =\int e^{f}f\relax (x) \frac {df}{dx}dx+C\\ y\relax (x) & =e^{-f}\int e^{f}fdf+e^{-f}C \end {align*}
But \(\int e^{f}fdf\) is the same as \(\int e^{x}xdx\) which by integration by parts gives \(e^{x}\left (x-1\right ) \) or in terms of \(f\), gives \(e^{f}\left (f-1\right ) \). Hence the above becomes\begin {align*} y\relax (x) & =e^{-f}\left (e^{f}\left (f-1\right ) \right ) +e^{-f}C\\ & =f-1+e^{-f}C \end {align*}