Take a look at this differential equation:
$$ (x^3 + xy^2-y)dx + (y^3+yx^2+x)dy = 0$$
For equation to be exact, we need to prove,
$$ \frac {\partial M}{\partial y} = \frac {\partial N}{\partial x} $$
but here,
$ M = x^3+xy^2-y $ and $\frac {\partial M}{\partial y} =2xy-1$
and,
$ N = y^3+yx^2+x $ and $\frac {\partial N}{\partial x} =2xy+1$
Clearly, $\frac {\partial M}{\partial y} \neq \frac {\partial N}{\partial x}$
But now, if we rearrange the differential equation like this, $$ xdx + ydy + \frac{xdy-ydx}{x^2+y^2} = 0$$
then we get, $\frac {\partial M}{\partial y} = \frac {\partial N}{\partial x}$ = $\frac {y^2-x^2}{(x^2+y^2)^2}$
Now I know this differential equation IS exact, as i have taken this from an example in a book, but to solve this exact differential equation, they have rearranged it first to the form I have mentioned above. Why can't we prove $\frac {\partial M}{\partial y} = \frac {\partial N}{\partial x}$ in its initial form?