They ask you to solve the following differential equation, taking as data its integrating factor $u(x,y)$:
$$(y+\sin x \cos^2(xy)-2x^2y)dx+(x+\sin y \cos^2(xy))dy=0; \quad u(x,y)=u(xy)$$
This is my advance, then I could not conclude something else, since I could not simplify because of the integrating factor they ask for.
$Sol:$
As:
$N\frac{\partial Lnu}{\partial x}-M\frac{\partial Lnu}{\partial y}=\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}$....(1)
$z=xy \rightarrow (\frac{\partial z}{\partial x}=y \wedge \frac{\partial z}{\partial y}=x)$
$\frac{\partial Lnu}{\partial x}=\frac{\partial Lnu}{\partial z}.\frac{\partial z}{\partial x}=\frac{\partial Lnu}{\partial z}.y$
$\frac{\partial Lnu}{\partial y}=\frac{\partial Lnu}{\partial z}.\frac{\partial z}{\partial y}=\frac{\partial Lnu}{\partial z}.x$
Replacing in (1):
$N(\frac{\partial Lnu}{\partial z}.y)-M(\frac{\partial Lnu}{\partial z}x)=\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}$
$\frac{\partial Lnu}{\partial z}=\frac{\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}}{yN-xM}....(2)$
As:
$M=y+\sin(x)\cos^2(xy)-2x^2y$
$N=x+\sin(y)\cos^2(xy)$
$M_y=1-2x\sin(y)\cos(xy)\sin(xy)-2x^2$
$N_x=1-2y\sin(y)\cos(xy)\sin(xy)$
Replacing in (2):
$\frac{\partial Lnu}{\partial z}=\frac{1-2x\sin \left(x\right)\cos \left(xy\right)\sin \left(xy\right)-2x^2-\left(1-2y\sin \left(y\right)\cos \left(xy\right)\sin \left(xy\right)\right)}{\left[y\left(x+\sin \left(y\right)\cos ^2\left(xy\right)\right)-x\left(y+\sin \left(x\right)\cos ^2\left(xy\right)-2x^2y\right)\right]}$
I cannot simplify this last expression, or give it the form to replace xy = z