19. Verify that $f_{x y}=f_{y x}$ for the following function.
\[
\begin{array}{l}
f(x, y)=2 \cos x y \\
f_{x}=\square \\
f_{y}=\square \\
f_{x y}=f_{y x}=\square
\end{array}
\]

Step 1 ：Given the function \(f(x, y)=2 \cos(xy)\).

Step 2 ：Find the first partial derivatives \(f_x\) and \(f_y\). The first partial derivative of \(f\) with respect to \(x\) is \(f_x = -2y\sin(xy)\) and with respect to \(y\) is \(f_y = -2x\sin(xy)\).

Step 3 ：Find the second partial derivatives \(f_{xy}\) and \(f_{yx}\). The second partial derivative of \(f\) with respect to \(x\) then \(y\) is \(f_{xy} = -2xy\cos(xy) - 2\sin(xy)\) and with respect to \(y\) then \(x\) is \(f_{yx} = -2xy\cos(xy) - 2\sin(xy)\).

Step 4 ：Compare \(f_{xy}\) and \(f_{yx}\) to see if they are equal. Since \(f_{xy} = f_{yx} = -2xy\cos(xy) - 2\sin(xy)\), the function \(f(x, y)=2 \cos(xy)\) satisfies the equality of mixed partials.

Step 5 ：\(\boxed{f_{xy}=f_{yx}=-2xy\cos(xy) - 2\sin(xy)}\) is the final answer.