\[ y^{\prime }=-\frac {x+y}{3x+3y-4}\] Comparing to \(y^{\prime }=\frac {a_{1}x+b_{1}y+c_{1}}{a_{2}x+b_{2}y+c_{2}}\) shows that \(a_{1}=-1,b_{1}=-1,a_{2}=3,b_{2}=3\). Hence \(\frac {a_{1}}{b_{1}}=1,\frac {a_{2}}{b_{2}}=1\). This shows the lines are parallel. Let
\begin {align*} U\left ( x\right ) & =a_{1}x+b_{1}y\\ & =-x-y \end {align*}
Hence \(y^{\prime }=-1-U^{\prime }\left ( x\right ) \). Hence the ode becomes\begin {align*} -1-U^{\prime }-\frac {U}{-3U-4} & =0\\ U^{\prime } & =-\frac {2U+4}{3U+4} \end {align*}
This is separable. After solving for \(U\left ( x\right ) \), then \(y\) is found from \(U\left ( x\right ) =-x-y\)\[ y=-x-U \]