Here $M= x^2-4xy-2y^2 $ & $ N= y^2-4xy-2x^2 $

$\dfrac {\partial M}{\partial y}=0-4x-4y$

$\dfrac {\partial N}{\partial x}=0-4y-4x$

Here,

$\dfrac {\partial M}{\partial y}=\dfrac {\partial N}{\partial x}$

Thus the equation is exact

$\therefore \int M dx=\int x^2-4xy-2y^2 dx$

$=\dfrac {x^3}3-\dfrac{4x^2}2 y -2y^2x$

And $\int$ terms in N free of $xdy =\int y^2 dy$

$=y^3/3$

$\therefore $ The solution is

$\dfrac {x^3}3-2x^2y-2xy^2 +\dfrac {y^3}3=c $