When doing implicit differentiation, if differentiating #y# with respect to #x# then #x# can be differentiated as usual (i.e. #d/dx(x) = 1#).

But #y# is a bit different. The derivative of #y# alone will not be #1# but #dy/dx#.

For any power or function of y the chain rule will have to be implemented. For example, the derivative of #y^3# will be #3y^2dy/dx#. To differentiate #-1=xy^3-x^2y# simply differentiate term by term. So:

#d/dx(-1)=0#

#d/dx(xy^3)=y^3+3xy^2dy/dx# (Don't forget the product rule!)

#d/dx(-x^2y) = -2xy-x^2dy/dx#

So putting these back into the given function; differentiating #-1=xy^3-x^2y# will give us:

#0=y^3+3xy^2dy/dx-2xy-x^2dy/dx#

Rearrange this to get dy/dx on one side of the equation:

#0 = y^3-2xy+(3xy^2-x^2)dy/dx#

#dy/dx(3xy^2-x^2)=2xy - y^3#

#dy/dx = (2xy - y^3)/(3xy^2-x^2)#

And, thus our final solution.