Every real-holomorphic Hamiltonian vector field on a Kähler manifold is Killing (and preserves curvature), yes? Following the notation of Moroianu's Lectures on Kähler Geometry, we let $(M,g,\Omega,J)$ denote the metric $g$, symplectic form $\Omega$, and complex structure $J$ of a Kähler manifold $M$, satisfying the compatibility condition $g(X,Y) = g(JX,JY) = \Omega(X,JY)$ for vector fields $X,Y$.  
Supposing further that $V\lrcorner\ \Omega = dH$ for $H$ a real-valued (explicitly biholomorphic) Hamiltonian potential $H:M\to\mathbb R$, such that V is real-holomorphic (is this correct?) then we immediately have the following Lie derivative relations: 
$$ \mathcal{L}_V \Omega = \mathcal{L}_V J = 0\quad\Rightarrow\quad  \mathcal{L}_V g = 0 $$
or equivalently this proposition: 


Proposition Every real-holomorphic Hamiltonian vector field on a Kähler manifold is Killing.


This proposition is (in essence) a Hamiltonian converse to the following proposition of Moroianu's 


Proposition 9.5 (Moroianu) Every Killing vector field on a compact Kähler manifold is real holomorphic.


Three specific questions are asked:
Q1  Is the proposition $\mathcal{L}_V g = 0$ correct (for the assumptions given)?
Q2  Where can it be found in the literature?
Q3  Does it "trivially" imply $\mathcal{L}_V \mathcal{R} = 0$, where $\mathcal{R}$ is the scalar curvature?
Note: My numerical calculations suggest $\mathcal{L}_V \mathcal{R} \ne 0$.  The practical question is simply which is buggy: the formal reasoning associated to the proposition, or the code, or the "trivial" expectation that Q3's answer is "yes"?  
Deficiencies in my understanding of terms like "real-holomorphic" are plausible (and even likely).  Answers/references/general advice are very welcome!
 A: If $\mathcal{L}_v J= 0$ and  $\mathcal{L}_v \Omega= 0$, then $\mathcal{L}_v g=0$ so $v$ is a Killing vector field.  Indeed, the property 
$\mathcal{L}_v J= 0$  is equivalent to the condition that the (local) 
 flow of $v$ preserves $J$  and the property  $\mathcal{L}_v \Omega= 0$ 
is equivalent to the condition that the (local)   flow of $v$ preserves $\Omega$. Since $g$ is costructable by $J$ and $\Omega$, if both  $J$ and $\Omega$ are preserved by the flow,  then  $g$ is preserved by the flow as  well and the vector field is Killing. 
Hamiltonian vector fields have the property    $\mathcal{L}_v \Omega= 0$. If your definition of ``real-holomorphic'' implies  $\mathcal{L}_v J= 0$, then the answer on your question is yes, yes, yes, ya, ya, ya. 
A: Late to the party, but here's some extra nugget of information for the record. If $(M,\omega, J,g)$ is Kähler, $\omega(X,Y) = g(JX,Y)$ implies that $$(\mathcal{L}_X\omega)(Y,Z) = (\mathcal{L}_Xg)(Y,Z) + g((\mathcal{L}_XJ)Y,Z),$$for all fields $X,Y,Z \in \mathfrak{X}(M)$. It follows from this that any two of the following conditions together imply the remaining one: $\mathcal{L}_X\omega = 0$, $\mathcal{L}_Xg = 0$, $\mathcal{L}_XJ = 0$. In other words, any two of the following conditions together imply the remaining one: $X$ is symplectic, $X$ is Killing, $X$ is real-holomorphic.
If we also assume, in addition, that $M$ is compact, then every Killing field is Hamiltonian and real-holomorphic, but while $X$ real-holomorphic implies $JX$ real-holomorphic, $X$ and $JX$ are not both Killing unless $X$ is parallel (with respect to the Levi-Civita connection of $g$). See the second chapter of Einstein Manifolds by Arthur Besse for more details.
