Functional equation for $\zeta_K(s)$
Let $w =i\sqrt{5}, K = \mathbb{Q}(w),\mathcal{O}_K = \mathbb{Z}[w]$. We'll prove the functional equation for the Dedekind zeta function $\zeta_K(s)$ by looking at each of its ideal class separately.
The ideal class group has two elements $C_0 = (1),C_1 = (2,1+w)$. Both are rank $2$ free $\mathbb{Z}$ modules :
$$(1) = \{ n+mw, (n,m) \in \mathbb{Z}^2\}, \qquad (2,1+w) = \{ 2n+m(1+w), (n,m) \in \mathbb{Z}^2\} \tag{eq.1}$$
There are $4$ units in $\mathcal{O}_K$
$$\zeta_K(s) = \sum_{I \subset \mathcal{O}_K} N(I)^{-s}=\sum_{I \in C_0} N(I)^{-s}+\sum_{I \in C_1} N(I)^{-s} \tag{eq.2}$$
$$= \frac14\sum_{(n,m) \ne (0,0)} N(n+mw)^{-s}+ \frac{2^s}4 \sum_{(n,m) \ne (0,0)} N(2n+m(1+w))^{-s} \tag{eq.3}$$
Where $2^s$ in front of the last sum is because we are looking at the ideal $\frac{2n+m(1+w)}{2} C_1$.
$N(n+mw) = n^2+5m^2$ and $N(2n+m(1+w)) = (2n+m)^2+5m^2$.
The 1st sum is easy. Let $$\theta(x) = \sum_{n=-\infty}^\infty e^{-\pi n^2 x}, \qquad \theta(1/x) =x^{1/2}\theta(x) \tag{eq.4}$$
$$\Lambda_{C_0}(s) = 5^{s/2}\pi^{-s}\Gamma(s)4 \sum_{I \in C_0} N(I)^{-s} = \int_0^\infty x^{s-1} (\theta(x5^{-1/2})\theta(x5^{1/2})-1)dx \tag{eq.5}$$
$$\int_0^1 x^{s-1} (\theta(x5^{-1/2})\theta(x5^{1/2})-1)dx=\frac{1}{-s}+\int_1^\infty x^{-s-1} \theta(5^{-1/2}/x)\theta(5^{1/2}/x)dx \tag{eq.6}$$
$$ =\frac{1}{-s}+\int_1^\infty x^{-s} \theta(x5^{1/2})\theta(x5^{-1/2})dx
=\frac{1}{s-1}+\frac{1}{-s}+\int_1^\infty x^{-s} (\theta(x5^{1/2})\theta(x5^{-1/2})-1)dx \tag{eq.7}$$
$$\Lambda_{C_0}(s) = \frac{1}{s-1}+\frac{1}{-s}+\int_1^\infty (x^{s-1}+x^{-s}) (\theta(x5^{1/2})\theta(x5^{-1/2})-1)dx= \Lambda_{C_0}(1-s) \tag{eq.8}$$
For treating $C_1$ we'll need some linear algebra $$A = {\scriptstyle\begin{pmatrix} 2 & 1 \\ 0 & \sqrt{5}\end{pmatrix}}, \qquad A^{-1} = \frac{1}{2\sqrt{5}} {\scriptstyle\begin{pmatrix} \sqrt{5} & -1 \\ 0 & 2\end{pmatrix}} \tag{eq.9}$$
The Fourier transform does transform nicely under matrix transformations :
$$\mathcal{F}[g (A u)](\omega) = \int g(Au)e^{-i\omega^T u}du=\int g(u)e^{-i\omega^T A^{-1}v}dA^{-1}v= \frac{1}{|\det(A)|}\mathcal{F}[g ( u)](A^{-T}\omega ) \tag{eq.10}$$
Let $g(a,b) = e^{-\pi (a^2+b^2)}$ which is its own Fourier transform. By the Poisson summation formula, with $x$ a constant :
$$\Theta(x) = \sum_{n \in \mathbb{Z}^2} g(x^{1/2} A n) = \frac{1}{x}\sum_{n \in \mathbb{Z}^2} \mathcal{F}[g(A u)](x^{-1/2}n)=\frac{1}{x|\det(A)|}\sum_{n \in \mathbb{Z}^2} \mathcal{F}[g( u)](x^{-1/2}A^{-T}n)\\ =\frac{1}{2x\sqrt{5}}\sum_{n \in \mathbb{Z}^2} g( x^{-1/2}A^{-T}n) = \frac{1}{2x\sqrt{5}}\sum_{n \in \mathbb{Z}^2} g( \frac{x^{-1/2}}{2 \sqrt{5}}An) = \frac{1}{2x\sqrt{5}}\Theta(\frac{1}{20 x})\tag{eq.11}$$
$$\vartheta(x) = \Theta(\frac{x}{\sqrt{20}}), \qquad \vartheta(x) = \frac{1}{x}\Theta(\frac{1}{\sqrt{20} x})=\vartheta(1/x)\tag{eq.12}$$
$$\Lambda_{C_1}(s) = 20^{s/2}\pi^{-s}\Gamma(s)4\sum_{I \in C_1} N(I)^{-s} = \int_0^\infty x^{s-1} \sum_{(n,m) \ne (0,0)} e^{-\pi ((2n+m)^2+5m^2)\frac{x}{\sqrt{20}}}dx=\int_0^\infty x^{s-1} (\vartheta(x)-1)dx \tag{eq.13}$$
$$\int_0^1 x^{s-1} (\vartheta(x)-1)dx=\frac{1}{-s}+ \int_1^\infty x^{-s-1} \vartheta(1/x)dx=\frac{1}{-s}+\int_1^\infty x^{-s} \vartheta(x)dx=\\ \frac{1}{s-1}+\frac{1}{-s}+\int_1^\infty x^{-s} (\vartheta(x)-1)dx \tag{eq.14}$$
$$\Lambda_{C_1}(s) = \frac{1}{s-1}+\frac{1}{-s}+\int_1^\infty (x^{s-1}+x^{-s}) (\vartheta(x)-1)dx= \Lambda_{C_1}(1-s) \tag{eq.14}$$
$$\Lambda(s)= 5^{s/2}\pi^{-s}\Gamma(s)4 \zeta_K(s) = 5^{s/2}\pi^{-s}\Gamma(s)4 (\sum_{I \in C_0} N(I)^{-s}+2^s \sum_{I \in C_1} N(I)^{-s})\\ =\Lambda_{C_0}(s)+\Lambda_{C_1}(s) = \Lambda(1-s) \tag{eq.15}$$