How to derive the standard euler equation from the Expanding Variety Model
The way you would go about solving this problem is as the ChinG said is by setting up the Hamiltonian. In this case this is:
$$\mathcal{H}:e^{-\rho t} \frac{C(t)^{1-\theta}-1}{1-\theta}+\mu(t)\left[Y(t)-C(t)-X(t)-Z(t)\right]$$
Taking the first order condition for this problem we get:
$$\frac{\partial \mathcal{H}}{\partial C(t)}:e^{-\rho t}C(t)^{-\theta}-\mu(t)=0 $$
or
$$e^{-\rho t}C(t)^{-\theta}=\mu(t)$$
Taking the natural logs of this equation and differentiating with respect to time we get:
$$\ln\left[\frac{\partial \mathcal{H}}{\partial C(t)}\right]:-\rho t -\theta \ln [C(t)]=\ln[\mu(t)]$$
$$\frac{\partial\ln\left[\mathcal{H}_{C(t)}\right]}{\partial t}:-\rho-\theta\frac{\dot{C}(t)}{C(t)}=-\frac{\dot{\mu}(t)}{\mu(t)}$$
Rearranging the above and noting that $\frac{\dot{\mu}(t)}{\mu(t)}=r(t)$
we get:
$$\frac{\dot{C}(t)}{C(t)}=\frac{1}{\theta}(r(t)-\rho)$$
Which is the Euler equation.
Hope this helps.