How to calculate the gradient of $f(x).$
$f(x) = \dfrac{1}{2} \|C\|^2$
$C = A\,\,{.\!^*}\,B$ .... element-wise multiplication
Volume $A$ has the dimensions $(x,y,z)=(3,3,3)$ and $B$ has the same size as $A.$ The element-wise product $C$ has the same size as $A$ and $B$.
$\dfrac{\partial f(x)}{\partial C} = \left(\dfrac{\partial f(x)}{\partial A},\dfrac{\partial f(x)}{\partial B}\right)$
I am using the chain rule:
$\dfrac{\partial f(x)}{\partial A} = \dfrac{\partial f(x)}{\partial C} \dfrac{\partial C}{\partial A} = \|C\| B$
$\dfrac{\partial f(x)}{\partial A} = \dfrac{\partial f(x)}{\partial C} \dfrac{\partial C}{\partial B} = \|C\| A$
And now the Euclidean norm must give a scalar, which is multiplied by $B$ or $A?$
It can be helpful in these sorts of situations to write out what you've got: \begin{align*} f(x)&=\frac12 \|C\|^2=\frac12 (C_x^2+C_y^2+C_z^2)\\ &=\frac12 [(A_xB_x)^2+(A_yB_y)^2+(A_zB_z)^2]. \end{align*} Now, it's not clear to me that $$\dfrac{\partial f(x)}{\partial C} = \left(\dfrac{\partial f(x)}{\partial A},\dfrac{\partial f(x)}{\partial B}\right)$$ is correct. Wouldn't it rather be $$\dfrac{\partial f(x)}{\partial C} = \left(\dfrac{\partial f(x)}{\partial C_x},\dfrac{\partial f(x)}{\partial C_y},\dfrac{\partial f(x)}{\partial C_z}\right)?$$ In that case, you'd have $$\dfrac{\partial f(x)}{\partial C} = \left(\dfrac{\partial f(x)}{\partial \|C\|}\dfrac{\partial \|C\|}{\partial C_x},\;\dfrac{\partial f(x)}{\partial \|C\|}\dfrac{\partial \|C\|}{\partial C_y},\;\dfrac{\partial f(x)}{\partial \|C\|}\dfrac{\partial \|C\|}{\partial C_z}\right),$$ or you could even do (which would be easier): \begin{align*}\dfrac{\partial f(x)}{\partial C} &= \left(\dfrac{\partial f(x)}{\partial (\|C\|^2)}\dfrac{\partial (\|C\|^2)}{\partial C_x},\;\dfrac{\partial f(x)}{\partial (\|C\|^2)}\dfrac{\partial (\|C\|^2)}{\partial C_y},\;\dfrac{\partial f(x)}{\partial (\|C\|^2)}\dfrac{\partial (\|C\|^2)}{\partial C_z}\right)\\&=\left(\|C\|C_x,\|C\|C_y,\|C\|C_z\right) \\ &=\|C\|(C_x,C_y,C_z)\\ &=\|C\|C. \end{align*} Not sure how $A$ and $B$ necessarily need to go into the problem.