We have, $\frac{d y}{d x}=\frac{x+7 y+3}{3 x+5 y+9}$
Put, $x=x+h$ and $y=y+k$
$\therefore \quad \frac{d y}{d x}=\frac{(x+7 y)+(h+7 k+3)}{(3 x+5 y)+(3 h+5 k+9)}$
Put $h+7 k+3=0$ and $3 h+5 k+9=0$
$\Rightarrow \quad h=-3, k=0$
$\therefore \quad \frac{d y}{d x}=\frac{x+7 y}{3 x+5 y}$
Now, put $y=v x \Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}$
$\Rightarrow \quad x \frac{d v}{d x}=\frac{1+7 v-3 v-5 v^2}{3+5 v}$
$\begin{aligned} & \Rightarrow \quad x \frac{d v}{d x}=-\left[\frac{5 v^2-4 v-1}{5 v+3}\right] \\ & \Rightarrow \int \frac{5 v+3}{5 v^2-4 v-1} d v=\int-\frac{1}{x} d x \\ & \Rightarrow \frac{1}{2} \int \frac{10 v+6}{5 v^2-4 v-1} d v=-\int \frac{1}{x} d x \\ & \Rightarrow \frac{1}{2} \int \frac{10 v-4+10}{5 v^2-4 v-1} d v=-\int \frac{1}{x} d x \\ & \Rightarrow \frac{1}{2} \int \frac{10 v-4}{5 v^2-4 v-1} d v+5 \int \frac{1}{5 v^2-4 v-1} d v \\ & =-\log x+\log k \\ & \Rightarrow \frac{1}{2} \log \left[5 v^2-4 v-1\right]+\frac{5}{5} \int \frac{1}{v^2-\frac{4}{5} v-\frac{1}{5}} d v=\log \left(\frac{k}{x}\right) \\ & \Rightarrow \frac{1}{2} \log \left[5 v^2-4 v-1\right]+\int \frac{1}{\left(v-\frac{2}{5}\right)^2-\frac{1}{5}-\frac{4}{25}} d v=\log \frac{k}{x} \\ & \Rightarrow \frac{1}{2} \log \left[5 v^2-4 v-1\right]+\int \frac{1}{\left(v-\frac{2}{5}\right)^2-\left(\frac{3}{5}\right)^2} d v=\log \frac{k}{x} \\ & \Rightarrow \frac{1}{2} \log \left[5 v^2-4 v-1\right]+\frac{1}{2 \times \frac{3}{5}} \log \frac{v-\frac{2}{5}-\frac{3}{5}}{v-\frac{2}{5}+\frac{3}{5}}=\log \frac{k}{x} \\ & \Rightarrow \frac{1}{2} \log \left[5 v^2-4 v-1\right]+\frac{5}{6} \log \frac{5 v-5}{5 v+1}=\log \frac{k}{x} \\ & \end{aligned}$
$\begin{aligned} & \Rightarrow \frac{1}{6}\left[\log \left(5 v^2-4 v-1\right)^3+\log \left(\frac{5 v-5}{5 v+1}\right)^5\right]=\log \frac{k}{x} \\ & \Rightarrow \log \left[\left(5 v^2-4 v-1\right)^3 \times 5^5 \frac{(v-1)^5}{(5 v+1)^5}\right]=\log \frac{k^6}{x^6} \\ & \Rightarrow(5 v+1)^3(v-1)^3 \times \frac{(v-1)^5}{(5 v+1)^5}=\frac{c}{x^6}, \text { where } c=\frac{k^6}{5^5} \\ & \Rightarrow \quad \frac{(v-1)^8}{(5 v+1)^2}=\frac{c}{x^6} \Rightarrow \frac{(v-1)^4}{5 v+1}=\frac{c}{x^3} \\ & \Rightarrow \quad \frac{(y-x)^4}{x^4\left[\frac{5 y+x}{x}\right]}=\frac{c}{x^3} \Rightarrow(y-x)^4=c(5 y+x) \\ & \Rightarrow \quad \frac{(y-x+3)^4}{2}=c(5 y+x-3)\end{aligned}$