Use the substitution \(z=x+y\) to find the solution of the differential equation
\(\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{1+3 x+3 y}{3 x+3 y-1}\)
for which \(y=0\) when \(x=1\). Give your answer in the form \(a \ln (x+y)+b(x-y)+c=0\), where \(a\), \(b\) and \(c\) are constants to be determined.