As $$ \frac{dx}{x}(δx−γ)=\frac{dy}{y}(α−βy) $$ there is indeed a first integral $$ F(x,y)=δx-γ\ln|x|+βy-α\ln|y| $$
Taking this as a Hamiltonian for a potentially different time scale results in the system \begin{align} \frac{dx}{ds}&=-F_y=\fracαy-β&&=\frac{αx-βxy}{xy}\\\ \frac{dy}{ds}&=~~~F_x=δ-\fracγx&&=\frac{δxy-γy}{xy}\\\ \end{align} so that for $xy>0$ the relation between the time scales is $\frac{dt}{ds}=\frac1{xy}$ To this Hamiltonian system you can apply the symplectic Euler method, which should yield orbits that are closer to closed than the simple application of Euler. With some more effort in the initial step one can easily modify this method to leapfrog Verlet, which has even better theoretical properties. To get not only the curve but the solution, you also need to integrate the equation for $t$.