$ M=\frac{y}{x}\sec y-\tan y$

$N=-(x-\sec y\log x)$

$\frac{\partial{M}}{\partial{y}}=\frac{1}{x}\sec y +\frac{y}{x}\sec y\tan y-\sec^2y$

$ \frac{\partial{N}}{\partial{x}}=-\left(1-\frac{1}{x}\sec y\right) $

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

The given Equation is not an exact differential equation.

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

$=\frac{-\left( 1-\frac{1}{x}\sec y\right)-\left( \frac{1}{x}\sec y+\frac{y}{x}\sec y\tan y-\sec^2y\right)}{\frac{y}{x}\sec y-\tan y}$

$= \frac{- 1+\frac{1}{x}\sec y-\frac{1}{x}\sec y-\frac{y}{x}\sec y\tan y+\sec^2y}{\frac{y}{x}\sec y-\tan y}$

$=\frac{- 1-\frac{y}{x}\sec y\tan y+\sec^2y}{\frac{y}{x}\sec y-\tan y}$

$=\frac{-\frac{y}{x}\sec y\tan y+\tan^2y}{\frac{y}{x}\sec y-\tan y}$

$=\frac{(-\frac{y}{x}\sec y+\tan y)\tan}{\frac{y}{x}\sec y-\tan y}=-\tan y=f(y)$

$I.F=e^{\int -\tan ydy}=e^{-\log\sec y}=\cos y$

Multiplying the given equation by the integrating factor, we get

$\left(\frac{y}{x}\sec y-\tan y\right)\cos ydx-(x-\sec y\log x)\cos ydy=0 \\ \left(\frac{y}{x}-s\sin x\right)dx-(x\cos y-\log x)dy=0\\ \frac{\partial{M}}{\partial{y}}=\frac{1}{x}-\cos y+\frac{1}{y}$

The given Equation is an exact differential equation Solution is given by:

$\int Mdx+\int (Terms\ in\ N\ not\ containing\ x)dy=c \\ y=constant\\ \int (\frac{y}{x}-\sin y)dx-\int 0dy=c\\ y\log x-x\sin y=c$