If y³ = x, how would you differentiate this with respect to x? There are three ways:

1) rewrite it as y = x to the power of 1/3 and differentiate as normal.
2) dx  =  3y²
    dy
Since dy  =      1  
        dx        dx/dy
dy  =   1
dx       3y²

3) Differentiate term by term and use the chain rule:
     y³   =     x
d (y³)  =   d (x)
dx           dx

The right hand side of this equation is 1, since the derivative of x is 1. However, to work out the left hand side we must use the chain rule.
The left hand side becomes:
d (y³) ×  dy
dy           dx
(although it is not correct to do so, at this level you can think of dy/dx as a fraction in the chain rule. In the line above, imagine that you can cancel the 'dy' s, leaving d/dx and y³, which is what we had in the previous line).
Therefore, 3y² × dy  = 1
                         dx
So   dy  =   1
      dx       3y²
In this example, method (2) is clearly the easiest. However, there are cases when the only possible method is (3).

Example:
Differentiate x² + y² = 3x, with respect to x.
d (x²) + d (y²)  =   d (3x)
dx          dx            dx
2x + d (y²) × dy  =  3
       dy          dx
2x + 2y dy  = 3
            dx
dy  =  3 - 2x
dx         2y