Separate the variables: \[\int\frac{y}{y+5}\,\mathrm{d}y=\int xe^{-3x}\,\mathrm{d}x.\]

Since \(y/(y+5)=1-5/(y+5)\), the left integral is \(y-5\ln\lvert y+5\rvert\). For the right integral, take \(u=x\) and \(\mathrm{d}v=e^{-3x}\mathrm{d}x\):

\[\int xe^{-3x}\,\mathrm{d}x=-\frac x3e^{-3x}+\frac13\int e^{-3x}\,\mathrm{d}x=-\frac x3e^{-3x}-\frac19e^{-3x}.\]

Therefore \[y-5\ln\lvert y+5\rvert=-\frac x3e^{-3x}-\frac19e^{-3x}+C.\]

At \(x=0,y=0\), \(-5\ln5=-1/9+C\), so \(C=1/9-5\ln5\). The branch containing the initial point has \(y+5>0\), hence

\[\boxed{y-5\ln\left(\frac{y+5}{5}\right)=\frac{1-(3x+1)e^{-3x}}9}.\]