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.

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}
