\[ y(x) f'(x)-f(x) f'(x)+y'(x)=0 \] ✓ Mathematica : cpu = 0.0132074 (sec), leaf count = 18
\[\left \{\left \{y(x)\to f(x)+c_1 e^{-f(x)}-1\right \}\right \}\] ✓ Maple : cpu = 0.011 (sec), leaf count = 15
\[ \left \{ y \left ( x \right ) =f \left ( x \right ) -1+{{\rm e}^{-f \left ( x \right ) }}{\it \_C1} \right \} \]
\begin {equation} \frac {dy}{dx}+y\left ( x\right ) \frac {df}{dx}=f\left ( x\right ) \frac {df}{dx} \tag {1} \end {equation}
Integrating factor \(\mu =e^{\int \frac {df}{dx}dx}=e^{f}\). Therefore (1) becomes\[ \frac {d}{dx}\left ( e^{f}y\left ( x\right ) \right ) =e^{f}f\left ( x\right ) \frac {df}{dx}\] Integrating\begin {align*} e^{f}y\left ( x\right ) & =\int e^{f}f\left ( x\right ) \frac {df}{dx}dx+C\\ y\left ( x\right ) & =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\left ( x\right ) & =e^{-f}\left ( e^{f}\left ( f-1\right ) \right ) +e^{-f}C\\ & =f-1+e^{-f}C \end {align*}