$$\frac{\mathrm d y}{\mathrm d x}=\frac{ax+by+c}{k(ax+by)+d}=\frac{ax+by+c}{k(ax+by+C)}$$ with $C=\frac{d}{k}$ is solvable with the change of function : $$Y(x)=ax+by+C \quad\to\quad y=\frac{Y-ax-C}{b}$$ $$\frac{\mathrm d Y}{\mathrm d x}-a=\frac{Y-C+c}{kY}=$$ $$\frac{\mathrm d Y}{\mathrm d x}=\frac{(1+ak)Y-C+c}{kY}$$ This is a separable ODE. $$x=\int \frac{kY}{(1+ak)Y-C+c}=k\frac{(1+ak)Y+(c-C)\ln|(1+ak)Y-C+c|}{(1+ak)^2}+\text{ constant}$$ The inverse function $Y(x)$ involves the Lambert's W function.