<s>[INST] For the differential equation given by $(x^2-y^2)dx+3xydy=0$ the general solution is $(x^2+2y^2)^3 =cx^2$ 
For the differential equation given by $$(x^2-y^2)dx+3xy \ dy=0$$ the general soultion is $$(x^2+2y^2)^3 =cx^2$$ Check this solution by differentiating and rewriting to get the original function.

I solved for $\frac{dy}{dx}$ and got $$\frac{dy}{dx}=\frac{cx-3x(x^2+2y^2)^2} {6y(x^2+2y^2)^2}$$ I was told to solve for $c$ and multiply by $dx$. but I cannot figure out how to get this to the original equation. My teacher said to use implicit differentiation so I solved for $\frac{dy}{dx}$.
 [/INST] Calling 
$$
f(x,y) = (x^2 + 2 y^2)^3 -c x^2 = 0
$$
we have
$$
df=f_x dx+f_y dy = 0\to y' = -\frac{f_x}{f_y} = \frac{x \left(c-3 \left(x^2+2 y^2\right)^2\right)}{6 y \left(x^2+2 y^2\right)^2}
$$
now substituting $c = \frac{\left(x^2+2 y^2\right)^3}{x^2}$ and simplifying we obtain
$$
y' = \frac{y}{3 x}-\frac{x}{3 y}
$$
as expected.
NOTE
now making $z = \frac yx \to y' = x z' + z$ we have the transformed DE
$$
x z' = \frac 13\left(z-\frac 1z\right)-z
$$
which is separable and can be easily solved.
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