<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] The hint says you should solve for $c$ before differentiating, so let's do that
$$ \frac{(x^2+2y^2)^3}{x^2} = c $$
Performing implicit differentiation on this form will eliminate $c$
$$ \frac{3x^2(x^2+2y^2)^2\left(2x + 4y\frac{dy}{dx}\right)-2x(x^2+2y^2)^3}{x^4} = 0 $$
Now you can solve for $\frac{dy}{dx}$ and finish the job
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