Differential equation: $(2x^2 + 3y^2 -7)x dx = (3x^2 + 2y^2 -8 )y dy $ 
$(2x^2 + 3y^2 -7)x dx = (3x^2 + 2y^2 -8 )y dy $

Attempt: 
After expanding, everything is neat except: $3x^2y dy - 3y^2 x dx$
I can't convert it to exact differential. 
Also, there's weird symmetry in the equation wrt the coefficients of  $x^2$ and $y^2$. But not sure how to utilise that symmetry. 
As an attempt, though, I reached this: 
$\dfrac{d(x^2 - y^2 -1)}{2(x^2- y^2 -1)}= \dfrac{y dy}{(2x^2 + 3y^2 -7)}$ which is not useful at all. 
Answer given is: 

 $(x^2 +y^2 - 3)=(x^2 + y^2 - 1)^5C$


Hint:
$$(2x^2 -4 + 3y^2 -3)x dx = (3x^2 -6+ 2y^2 -2 )y dy$$
and use substitutions $x^2-2=u$ and $y^2-1=v$. You will find a homogeneous differential equation.
