Find the values

We are given with an equation $4 x+3 y=\log (4 x-3 y)$, we have to find $\frac{d y}{d x}$ of it, so by differentiating the equation on both sides with respect to $x$, we get,

$4+3 \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{(4 \mathrm{x}-3 \mathrm{y})}\left[4-3 \frac{\mathrm{dy}}{\mathrm{dx}}\right]$

$3 \frac{d y}{d x}+\frac{3}{(4 x-3 y)} \frac{d y}{d x}=\frac{4}{(4 x-3 y)}-4$

$\frac{d y}{d x}\left[1+\frac{1}{4 x-3 y}\right]=\frac{12 y-16 x+4}{3(4 x-3 y)}$

$\frac{d y}{d x}=\frac{\frac{12 y-16 x+4}{3(4 x-3 y)}}{\frac{4 x-3 y+1}{4 x-3 y}}=\frac{12 y-16 x+4}{12 x-9 y+3}$