Consider the stochastic process \begin{align*} dX_t&=\mu_t dt+\sqrt{Y_t}X_t dB_t \\ dY_t&=m_t dt+s\sqrt{Y_t} dW_t \end{align*} where $dB_tdW_t=\rho dt$, $s$ is a constant and $\mu_t$ and $m_t$ are well-behaved drifts.
According to Ito's formula: \begin{align*} df(X_t,Y_t) &= \left(\frac{1}{2}Y_tX_t^2f_{xx}(X_t,Y_t)+\rho s Y_tX_t f_{xy}(X_t,Y_t)+\frac{1}{2}s^2Y_tf_{yy}(X_t,Y_t)+\mu_t f_x(X_t,Y_t)+m_t f_y(X_t,Y_t)\right)dt \\ &\;\;\;\;\; +\sqrt{Y_t} X_t f_{x}(X_t,Y_t)dB_t+s\sqrt{Y_t} f_{y}(X_t,Y_t)dW_t. \end{align*} My question: Is the following assertion correct? \begin{align*} df(X_t,Y_t)^2 &= \bigg( Y_t X_t^2f_{x}(X_t,Y_t)^2+2\rho s X_tY_t f_{x} (X_t,Y_t)f_y(X_t,Y_t)+s^2 Y_t f_y(X_t,Y_t)^2\bigg)\text{d}t \end{align*} This expression then gives $\operatorname{Var}[df]=E[df^2]-E[df]^2=E[df^2]=df^2$, because $E[df]$ is of order $dt$ and squaring it makes it negligible whereas $df^2$ is non-random.
Just to mark this question as answered:
@Lutz agrees that the equation in the question is correct \begin{align*} df(X_t,Y_t)^2 &= \bigg( Y_t X_t^2f_{x}(X_t,Y_t)^2+2\rho s X_tY_t f_{x} (X_t,Y_t)f_y(X_t,Y_t)+s^2 Y_t f_y(X_t,Y_t)^2\bigg)dt. \end{align*}