I struggle with taking the derivative of the Hadamard-Product?

Let us consider $f(x)=x^TAx=x^T(Ax)$. We know 

$$\frac{\partial}{\partial x} x^TAx = (A+A^T)x.$$

The Matrix-Cookbook claimed
$d(XY)=d(X)Y+Xd(Y)$ and
$$\frac{\partial}{\partial x} x^Ta = \frac{\partial}{\partial x}a^Tx = a.$$

Setting $X:=x^T$ and $Y:=Ax$ we have
\begin{align\*}
X &= x^TE & d\_x(X) = E\\
Y &= Ax & d\_x(Y) = A\\
\end{align\*}
This gives
$$d(XY)=d(X)Y+Xd(Y)= 1^TAx + x^TA= Ax + x^TA$$
This format (dimension) is incorrect.
A generic rule seems to be $d(XY)=d(X)Y+(Xd(Y))^T$ instead? 
What does a generic rule look like for the Hadamard product $[a\odot b]\_i=[a]\_i\cdot [b]\_i$? The Matrix-Cookbook states:
$$d(X\odot Y)=d(X)\odot Y+X\odot d(Y).$$

For example, the derivative of
$$ (x\odot y)^TA(x\odot y)$$
we have
\begin{align\*}
X &= (x\odot y)^TE & d\_x(X) &= \left(d(x\odot y)\right)^TE\\
& & &= \left(1\odot y + x\odot 0\right)^TE\\
& & &= y^TE\\
Y &= A(x\odot y) & d\_x(Y) &= A\left(d(x\odot y)\right)\\
 & & &=A\left(1\odot y + x\odot 0\right)\\
 & & &=Ay\\
\end{align\*}

Which implies 
\begin{align\*}
d\_x((x\odot y)^TA(x\odot y))&=y^T\odot A(x\odot y) + \left( (x\odot y)^T Ay \right)^T\\
&=y^T\odot A(x\odot y) + y^TA^T(x\odot y)
\end{align\*}
as a derivative. But the correct derivative should be

$$2y\odot A(x\odot y)$$

What is missing?