If $x^4+7x^2y^2+9y^4=24xy^3$,show that $\frac{dy}{dx}=\frac{y}{x}$ If $x^4+7x^2y^2+9y^4=24xy^3$,show that $\frac{dy}{dx}=\frac{y}{x}$
I tried to solve it.But i got stuck after some steps.
$x^4+7x^2y^2+9y^4=24xy^3$
$4x^3+7x^2.2y\frac{dy}{dx}+7y^2.2x+36y^3.\frac{dy}{dx}=24x.3y^2\frac{dy}{dx}+24y^3$
$\dfrac{dy}{dx}=\dfrac{24y^3-4x^3-14y^2x}{14x^2y+36y^3-72xy^2}$
How to move ahead?Or there is some other elegant way to solve it.
 A: Put $u=\dfrac{y}{x} \to 9u^4-24u^3+7u^2+1=0\to 36u^3u'-72u^2u'+14uu'=0\to 2u(18u^2-36u+7)u'=0$. This means $u =0, u'=0$ or $18u^2-36u+7=0$. You can check that the first and the last can't happen for $u$. Thus the second is true, this means $y'=\dfrac{y}{x}$. 
A: Notice, we have $$x^4+7x^2y^2+9y^4=24xy^3\tag 1$$
Now, differentiating both the sides w.r.t. $x$ as follows $$\frac{d}{dx}(x^4+7x^2y^2+9y^4)=\frac{d}{dx}(24xy^3)$$
$$4x^3+14x^2y\frac{dy}{dx}+14xy^2+36y^3\frac{dy}{dx}=72xy^2\frac{dy}{dx}+24y^3$$
$$\frac{dy}{dx}(14x^2y+36y^3-72xy^2)=24y^3-4x^3-14xy^2$$ 
$$\frac{dy}{dx}=\frac{24y^3-4x^3-14xy^2}{14x^2y+36y^3-72xy^2}$$
$$=\frac{y}{x}\cdot  \frac{x}{y}\left(\frac{24y^3-4x^3-14xy^2}{14x^2y+36y^3-72xy^2}\right)$$
$$=\frac{y}{x}\left(\frac{24xy^3-4x^4-14x^2y^2}{14x^2y^2+36y^4-72xy^3}\right)$$
Now, substituting the value of $x^4$ from (1), we get
 $$\frac{dy}{dx}=\frac{y}{x}\left(\frac{24xy^3-4(24xy^3-7x^2y^2-9y^4)-14x^2y^2}{14x^2y^2+36y^4-72xy^3}\right)$$
$$=\frac{y}{x}\left(\frac{14x^2y^2+36y^4-72xy^3}{14x^2y^2+36y^4-72xy^3}\right)=\frac{y}{x}$$ Hence, we get 
$$\bbox[5pt, border:2.5pt solid #FF0000]{\color{blue}{\frac{dy}{dx}=\frac{y}{x}}}$$
A: $$x^4+7x^2y^2+9y^4=24xy^3$$
Can be rewritten as,
$$1+7\frac{y^2}{x^2}+9 \frac{y^4}{x^4}=24\frac{y^3}{x^3}$$
Differentiate wrt $x$, you will get that three things can be equated to $0$. One of them will lead to your result.
(Hint: Don't expand derivative using quotient rule.)
A: Try computing $$\frac xy \frac {dy}{dx}$$ and then using the original identity to substitute terms in $xy^3$ in numerator and denominator.
A: Another way to view this is that the differential equation $ \ \frac{dy}{dx} \ = \ \frac{y}{x} \ $ is separable and has the general solution given by
$$ \ \int \ \frac{dy}{y} \ = \ \int \ \frac{dx}{x} \ \ \Rightarrow \ \ \ln |y| \ = \ \ln |x| \ + \ C \ \ \Rightarrow \ \ y \ = \ A \ x \ \ , $$
with $ \ A \ $ being any real number.  This is reasonable, since the differential equation can be interpreted as saying that the slope of the solution curve(s) at any point is   $ \ \frac{y}{x} \ $ . The origin $ \ ( 0 , \ 0 ) \ $ is a point on the curve of the given equation, so the slope of the curve everywhere must be constant.
We can solve for the constant(s) by inserting our general solution into the curve equation to obtain
$$x^4 \ + \ 7x^2y^2 \ + \ 9y^4 \ - \ 24xy^3 \ = \ 0 \ \ \rightarrow \ \ x^4 \ + \ 7x^2 \ (Ax)^2 \ + \ 9 \ (Ax)^4 \ - \ 24x \ (Ax)^3 \ = \ 0 $$
$$ \Rightarrow \ \ x^4 \ ( \ 9A^4 \ - \ 24A^3 \  + \ 7 A^2 \ + 1 \ ) \ = \ 0 \ \ .  $$
Other than at the origin ( $ \ x = 0 \ $ ) , the curve has two real slopes, $ \ A \ \approx \ 2.3230 \ $ and $ \ A \ \approx \ 0.54349 \ $ (the other two solutions for $ \ A \ $ are a complex conjugate pair).  If we plot the equation  $ \ x^4 \ + \ 7x^2y^2 \ + \ 9y^4 \ - \ 24xy^3 \ = \ 0 \ $ , we find that we in fact have a degenerate quartic curve, appearing as a pair of straight lines (the "wiggles" near the origin are a graphing artifact, as they are present at all plotting scales in the same way).

