$ \DeclareMathOperator{\cov}{cov} \DeclareMathOperator{\E}{\mathbb{E}} \renewcommand{\endproof}{\quad_\square} $ Suppose $X \sim \mathcal{N}(\mu_X, \Sigma_{XX})$, $Y \sim \mathcal{N}(\mu_X, \Sigma_{YY})$, $\cov[X,Y] = \Sigma_{XY}$.
Is it true that $\E_{X,Y}[X \cdot \sin{Y}] = \E_Y[\E_{X|Y}[X|Y]\cdot \sin{Y}]$? What about the more general case, $\E_{X,Y}[X \cdot f(Y)] = \E_Y[\E_{X|Y}[X|Y]\cdot f(Y)]$ for arbitrary $f$?
I can show that $\E_{X,Y}[X \cdot Y] = \E_Y[\E_{X|Y}[X|Y] \cdot Y]$:
It is a standard result that $\E_{X,Y}[X \cdot Y] = \Sigma_{XY} + \mu_X \cdot \mu_Y$. From this answer on stats.stackexchange.com, we have that $$\E_{X|Y}[X|Y] = \mu_X + \frac{\Sigma_{XY}}{\Sigma_{YY}}(Y - \mu_Y).$$ Now $$\begin{split} \E_Y[\E_{X|Y}[X|Y] \cdot Y] &= \E_Y[(\mu_X + \frac{\Sigma_{XY}}{\Sigma_{YY}}(Y - \mu_Y)) \cdot Y] \\ &= \E_Y[\mu_X \cdot Y] + \frac{\Sigma_{XY}}{\Sigma_{YY}}\E_Y[Y^2 - \mu_Y \cdot Y]\\ &=\mu_x \cdot \mu_Y + \frac{\Sigma_{XY}}{\Sigma_{YY}}(\E_Y[Y^2] - \E_y[Y]^2) \\ &=\mu_x \cdot \mu_Y + \frac{\Sigma_{XY}}{\Sigma_{YY}}\Sigma_{YY} \\ &=\Sigma_{XY} + \mu_X \cdot \mu_Y = E_{X,Y}[XY]. \endproof \end{split}.$$
I am also aware of the tower rule $\E_X[X] = \E_Y[\E_{X|Y}[X|Y]]$, and I can sketch a proof along the lines of
$$\begin{split} \E_{X,Y}[X \cdot f(Y)] &= \E_Y[\E_X[X \cdot f(Y)]] \\ &= \E_Y[\E_X[X] \cdot f(Y)] \\ &= \E_Y[\E_Y[\E_{X|Y}[X|Y]] \cdot f(Y)] \\ &= \E_Y[\E_{X|Y}[X|Y]] \cdot \E_Y[f(Y)] \\ &=\ \dots\ ? \end{split}$$
But this is clearly wrong, since $\E_{X,Y}[X\cdot Y] \neq \E_X[X]\cdot \E_Y[Y]$ as the above would imply.