You know that this integral should be zero by Cauchy's integral theorem but I suppose they want you to compute it. So here it goes:
Set $r := \frac{1 + \sqrt{5}}{2}$ and $\gamma (t) := r e^{i t}$. Then \oint f(z) dz = \int_0^{2 \pi} f(\gamma) \dot{\gamma}(t) dt = \int_0^{2 \pi} e^{4r (\cos t + i \sin t) + 1} r(i \cos t - \sin t) dt
Then $ \frac{d}{dt}\left ( e^{4r (\cos t + i \sin t) + 1} \right ) = e^{4r (\cos t + i \sin t) + 1} 4r(i \cos t - \sin t)$
So \oint f(z) dz = \left [ \frac{1}{4} e^{4r (\cos t + i \sin t) + 1} \right ]_0^{2 \pi} = \left [ \frac{1}{4} e^{4r e^{it} + 1} \right ]_0^{2 \pi} = 0