Suppose $E/\mathbb{Q}$ is an elliptic curve that has complex multiplication by $\mathcal{O}_K$, where $K=\mathbb{Q}(\sqrt{D})$, for $D<0$ and squarefree.
In "The main conjectures of Iwasawa theory for imaginary quadratic fields" by Karl Rubin, it is mentioned that the $L$-function of $E$ over $K$ satisfies
\begin{equation*} L(E/K,s) = L(E/\mathbb{Q},s)^2 \end{equation*}
I understand that in general, $L(E/K,s) = L(E/\mathbb{Q},s)\cdot L(E^D/\mathbb{Q},s)$, where $E^D/\mathbb{Q}$ is the quadratic twist of $E$. I'm guessing that the above equation holds because $E^D$ is $\mathbb{Q}$-isogenous to $E$, however I am unable to see why this is so, is there a way of showing this?
Are you sure it is written on Rubin's paper? Can you give a more precise reference? I fastly checked the paper but I did not find you assertion.
The formula (1) $$ L(E/K,s)=L(E/ \mathbb{Q})^2 $$
in general is false, unless you have some extra hypotesis you did not mention.
Indeed, assume that the analytic rank of $E/\mathbb{Q}$ is one.
By a result of Murty-Murty and Waldspurger there exists a quadratic imaginary field $K$ of negative discriminant (and some other properties), such that $\operatorname{ord}_{s=1}L(E/K,1)=1$.
Note that this contradicts the formula (1) , while it is compatible with the equality
$$ L(E/K,s) = L(E/\mathbb{Q},s)\cdot L(E^D/\mathbb{Q},s). $$ and in particular implies the non-vanishing of $L(E^D/\mathbb{Q},1)$.