<s>[INST] Find $y'$ given $y\,\sin\,x^3=x\,\sin\,y^3$? The problem is
$$y\,\sin\,x^3=x\,\sin\,y^3$$
Find the $y'$
The answer is

Can some explain how to do this, please help.
 [/INST] $y \sin x^3=x\sin y^3→$ Differentiating  both sides with respect to  $x .$
$⇒y(\cos x^3×3x^2)+y′\sin x^3 =\sin y^3+x \cos y^3×3y^2y′ $
$⇒3x^2y \cos x^3+y′\sin x^3=\sin y^3+3y^2x \cos y^3y′$
$⇒y′[\sin x^3−3y^2x\cos y^3]=\sin y^3−3x^2y \cos x^3$
$⇒y′=\dfrac{\sin y^3−3x^2y \cos x^3}{\sin x^3−3y^2x \cos y^3}$
I was searching it and I got the complete solution from here,
Implicit Differentiation
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