$ M=(xy\sin xy+\cos xy)ydx=xy^2 \sin xy-y\cos xyk] dy=0$

$N=(xy \sin xy-\cos xy)x=x^2y\sin xy-x\cos xy$

$\frac{\partial M}{\partial y}=2xy\sin xy+y^2x^2\cos xy+\cos xy=xy\sin xy$

$\frac{\partial N}{\partial x}=2xy \sin xy+x^2y^2\cos xy-\cos xy+xy\sin xy$

$\frac{\partial M}{\partial y}\neq\frac{\partial N}{\partial x}$

The given Equation is not an exact differential equation.

The given equation is of the form $f_1(xy)ydx+f_2(xy)xdy=0$

$I.F=\frac{1}{Mx-Ny}$

$ =\frac{1}{(xy^2\sin xy+y\cos xy)x-(x^2y\sin xy-x\cos xy)y}$

$=\frac{1}{(x^2y^2\sin xy+xy\cos xy)-(x^2y^2\sin xy-xy\cos xy)}$

$=\frac{1}{2xy\cos xy}$

Multiplying the given equation by the integrating factor, we get

$(xy^2\sin xy+y\cos xy)\frac{1}{2xy \cos xy}dx+(x^2y\sin xy-x\cos xy)\frac{1}{2xy \cos xy}dy=0$

$\left( \frac{y\sin xy}{2\cos xy}+\frac{1}{2x}\right)+\left( \frac{x\sin xy}{2\cos xy}-\frac{1}{2y}=\right)dy=0$

$\left( \frac{y\tan xy}{2}+\frac{1}{2x}\right)dx+\left( \frac{x\tan xy}{2}-\frac{1}{2y}\right)dy$

$\frac{\partial M}{\partial y}= \frac{xy\sec^2 xy+\tan xy}{2}$

$ \frac{\partial N}{\partial x}=\frac{xy\sec^2 xy +\tan xy}{2}$

The given Equation is an exact differential equation

Solution is given by

$\int Mdx+\int (Terms\ in\ N\ not\ containing\ x)dy=c$

$y=constant$

$\int \left( \frac{y \tan xy}{2}+\frac{1}{2x}\right)dx+ \int -\frac{1}{2y}dy=c$

$\int \frac{y\log \sec xy}{2y}+\frac{1}{2}\log x+\frac{1}{2}\log{y}=c$

$\frac{\log\sec(xy)}{2}+\frac{1}{2}\log x+\frac{1}{2}\log y=c $