I'm given the facts that $X$, $Y$ are jointly normal random vectors and the matrix $A$ solves $AC_{Y}=C_{XY}$, where $C_{Y}$ is the correlation matrix for $Y$ and $C_{XY}$ is the cross-correlation matrix between $X$ and $Y$. I try to proceed by computing the correlation matrix
$$ \begin{align} \text{Corr}\left(\left(X-m_{X}\right)-A\left(Y-m_{Y}\right),Y\right) &=E[{\left(\left(X-m_{X}\right)-A\left(Y-m_{Y}\right)\right)Y^{\intercal}} ]\\ &=E[{XY^{\intercal}-m_{X}Y^{\intercal}-AYY^{\intercal}+Am_{Y}Y^{\intercal}}] \\ &=E[{XY^{\intercal}}]-m_{X}E[{Y^{\intercal}}]-AE[{YY^{\intercal}}]+Am_{Y}E[{Y^{\intercal}}] \\ &=C_{XY}-m_{X}E[{Y^{\intercal}}]-AC_{Y}+Am_{Y}E{Y^{\intercal}} \\ &=\left(-m_{X}+Am_{Y}\right)m_{Y}^{\intercal} \end{align} $$
After this step I don't really know how to proceed. I think I would have to prove that $-m_{X}+Am_{Y}=0$ but I'm not sure how to go about showing that.
If $C_Y$ and $C_{XY}$ indeed are "correlation" and "cross-correlation" matrices as written in the OP, the claim is false.
For example, suppose that $X$ and $Y$ are scalar, with mean $0$ each, variances $4$ and $9$, respectively, and covariance $2$ between them. Then: $$ C_Y=1 \text{ (always}),\quad C_{XY}=\frac{2}{2\times 3}=\frac{1}{3},\quad A=\frac{1}{3}. $$ So with $Z\equiv (X-0)-\frac{1}{3}(Y-0)=X-\frac{1}{3}Y$, the covariance between $Z$ and $Y$ is $$ E(ZY)-E(Z)E(Y)=E(ZY)=E(XY)-\frac{1}{3}E(Y^2)=2-\frac{9}{3}\neq 0. $$
On the other hand, if $C_Y$ and $C_{XY}$ were "covariance" and "cross-covariance" matrices, then problem is straightforward: $$ \operatorname{Cov}[(X-m_X)-A(Y-m_Y),Y]=\underbrace{\operatorname{Cov}[X-m_X,Y]}_{C_{XY}}-A\underbrace{\operatorname{Cov}[Y-m_Y,Y]}_{C_Y}=0. $$