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.
 A: Differentiating with respect to $x$, we get :
$$y'\sin(x^3)+3x^2y\cos(x^3)=\sin(y^3)+3xy^2y'\cos(y^3)$$
At this point, we can easily compute $y'$ in terms of $x$ and $y$.
A: Rewrite this as $$y(x)\sin{x^3}=x\sin{y(x)^3}$$ to see that everything depends on $x$. Now, take the derivative with respect to $x$ (as you know) and solve for $y'(x)$. In the end rewrite again $y'(x)$ as $y'$ and $y(x)$ as $y$ to conform with the usual notation: 
\begin{align}0&=\frac{d}{dx}\left(y(x)\sin{x^3}-x\sin{y(x)^3}\right)\\[0.3cm]&=y'(x)\sin{x^3}+y(x)\cos{x^3}(x^3)'-\sin{y(x)^3}-x\cos{y(x)^3}(y(x)^3)'\\[0.2cm]&=y'(x)\sin{x^3}+3x^2y(x)\cos{x^3}-\sin{y(x)^3}-3xy(x)^2y'(x)\cos{y(x)^3}\\[0.3cm]\implies y'(x)&=\frac{\sin{y(x)^3}-3x^2y(x)\cos{x^3}}{\sin{x^3}-3xy(x)^2\cos{y(x)^3}}\\[0.3cm]\implies y'&=\frac{\sin{y^3}-3x^2y\cos{x^3}}{\sin{x^3}-3xy^2\cos{y^3}}\end{align} Now, you can repeat the same process by treating $x$ as a function of $y$. This will give you the second expression in the solution. But you do not need to do this.
A: $y \sin x^3=x\sin y^3→$ 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
A: I have found that it helps students to understand implicit differentiation if first they think of both $x$ and $y$ as functions of some third variable such as $t$ and take the derivative of both sides with respect to $t$, being careful to use the product rule, chain rule, etc when needed. Then as a final step, multiply both sides by $dt$. Since the object usually is to solve for $y^\prime=\frac{dy}{dx}$ you can then divide both sides by $dx$ and solve.
\begin{eqnarray}
\frac{d}{dt}\left[y\sin(x^3)\right]&=&\frac{d}{dt}\left[x\sin(y^3)\right]\\
\sin(x^3)\frac{dy}{dt}+y\frac{d\sin(x^3)}{dt}&=&\sin(y^3)\frac{dx}{dt}+x\frac{d\sin(y^3)}{dt}\\
\sin(x^3)\frac{dy}{dt}+3x^2y\cos(x^3)\frac{dx}{dt}&=&\sin(y^3)\frac{dx}{dt}+3xy^2\cos(y^3)\frac{dy}{dt}\\
\sin(x^3)dy+3x^2y\cos(x^3)dx&=&\sin(y^3)dx+3xy^2\cos(y^3)dy\\
\sin(x^3)\frac{dy}{dx}+3x^2y\cos(x^3)&=&\sin(y^3)+3xy^2\cos(y^3)\frac{dy}{dx}\\
\left[\sin(x^3)-3xy^2\cos(y^3)\right]\frac{dy}{dx}&=&\sin(y^3)-3x^2y\cos(x^3)\\
\frac{dy}{dx}&=&\frac{\sin(y^3)-3x^2y\cos(x^3)}{\sin(x^3)-3xy^2\cos(y^3)}
\end{eqnarray}
