<s>[INST] 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.
 [/INST] $$x^4+7x^2y^2+9y^4=24xy^3$$
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.)
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