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.
