Since this involves only powers of x and y I would look for an integrating factor of the form \(\displaystyle x^ny^m\). Multiplying by that we have \(\displaystyle Mdy+ Ndx= (2x^{m+4}y^n+ x^{m+1}y^{n+1})dy- 6x^my^{n+2}dx= 0\). Then \(\displaystyle M_x= 2(m+ 4)x^{m+3}y^n+ (m+1)x^my^{n+1}\) and \(\displaystyle N_y= -6(n+2)x^my^{n+1}\). This equation will be exact if and only if those are equal:\(\displaystyle M_x= 2(m+ 4)x^{m+3}y^n+ (m+1)x^my^{n+1}= -6(n+2)x^my^{n+1}\). Comparing powers of x, we see that we have \(\displaystyle x^my^{n+1}\) on both sides except for that first \(\displaystyle x^{m+3}y^n\). We can get rid of that by making the coefficient, 2(m+4), 0: we need m= -4. With m= -4, we must have \(\displaystyle -3x^{-4}y^{n+1}= -6(n+2)x^{-4}y^{n+1}\) so we need \(\displaystyle -3= -6(n+2)\) so \(\displaystyle n+ 2= \frac{1}{2}\) or \(\displaystyle n= -\frac{3}{2}\). An integrating factor is \(\displaystyle x^{-4}y^{-3/2}\).