<s>[INST] 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: 

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

 [/INST] Given $(2x^2+3y^2-7)x\ dx=(3x^2+2y^2-8)y\ dy$
Let us take $$X=x^2$$$$Y=y^2$$and we get$$\dfrac{dY}{dX}=\dfrac{2X+3Y-7}{3X+2Y-8}$$Again let us consider $$X=p+a$$$$Y=q+b$$$$\implies\dfrac{dq}{dp}=\dfrac{2p+2a+3q+3b-7}{3p+3a+2q+2b-8}=\dfrac{2p+3q}{3p+2q}$$From $2a+3b-7=0$ and $3a+2b-8=0$ we get $\implies a=2,b=1$
$$\dfrac{dq}{dp}=\dfrac{2p+3q}{3p+2q}\mbox{ is a homogeneous equation. So, the change of variable is given by}$$$$q(p)=p\ u(p)\implies u+s\dfrac{du}{dp}=\dfrac{2+3u}{3+2u}$$$$s\dfrac{du}{dp}=\dfrac{2+3u}{3+2u}-u=\dfrac{2-2u^2}{3+2u}$$$$\dfrac{dp}{p}=\dfrac{3+2u}{2-2u^2}du$$$$\ln|p|=\int\dfrac{3+2u}{2-2u^2}\ du=\dfrac14\ln|u+1|-\dfrac54-\ln|u-1|+C$$
$$s=C\dfrac{(u+1)^{\frac14}}{(u-1)^{\frac54}}\implies(q-p)^{\frac54}=C(q+p)^{\frac14}$$
$$(q-p)^5=C(q+p)\implies(q-p)^5-C(q+p)=0$$$$p=X-2\mbox{ }\mbox{ and }q=Y-1$$$$\implies(Y-X+1)^5-C(Y+X-3)=0$$Therefore,$$\boxed{(x^2 +y^2 - 3)=(x^2 + y^2 - 1)^5C}$$
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