According to the comments the only ugly part was the devectorization.
Consider the 3x3 matrix $${\bf X}=\left[\begin{array}{ccc}
1&2&3\\4&5&6\\7&8&9\end{array}\right]$$
It's lexical vectorization is:
$$\text{vec}({\bf X}) = \left[\begin{array}{ccccccccc}1&4&7&2&5&8&3&6&9\end{array}\right]^T$$
We consider that kind of vectorization here.
The Matlab / Octave expression:
kron([1,0,0]',eye(3))'*kron(R(:),[1,0,0])+
kron([0,1,0]',eye(3))'*kron(R(:),[0,1,0])+
kron([0,0,1]',eye(3))'*kron(R(:),[0,0,1])-R
evaluates to the zero matrix for several random matrices R, each new term "selecting" a new row.
Then a guess would be that
$${\bf R}=\sum_k({\bf v_k} \otimes {\bf I})^T(\text{vec}({\bf R})\otimes {\bf v_k}^T)$$
With $\bf v_k$ being the Dirac or selector vector:
$$({\bf v_k})_i = \cases{0, k \neq i\\1, k=i}$$
Feel free to try and simplify it if you want.
Let's denote the elementwise/Hadamard and inner/Frobenius products respectively as
$$\eqalign{
A &= B\circ C \cr
\alpha &= B:C = {\rm tr}(B^TC) \cr
}$$
Recall that these products are commutative, and mutually commutative
$$\eqalign{
A\circ B &= B\circ A \cr
A:B &= B:A \cr
A:B\circ C &= A\circ B:C \cr
}$$ and that the matrix of all ones is the identity element for the Hadamard product. Note that the matrices $(A,B,C)$ must have the same shape for these products to make sense.
Your scalar function can be written as
$$\eqalign{
y
&= 1:f \cr
&= 1:(Ax)\circ(Bx) \cr
&= Ax:Bx \cr
}$$
Whosee differential and gradient are
$$\eqalign{
dy
&= A\,dx:Bx + Ax:B\,dx \cr
&= Bx:A\,dx + Ax:B\,dx \cr
&= (A^TB + B^TA)\,x:dx \cr
\cr
\frac{\partial y}{\partial x}
&= (A^TB + B^TA)\,x \cr\cr
}$$
(Note that I've used juxtaposition rather than $*$ for the ordinary matrix product.)
Best Answer
No.
$B=\frac {1}{2}\begin{pmatrix}1+i&1-i\\1-i&1+i\end{pmatrix} $ is unitary .
It is your task to compute $A=B\circ B$ and prove that $\det(A)=0.$ Hence $A$ is not unitary.