Given:
f(x)=(sin^2x+cosx)/(sinx-cos^2x)
Let u=sin^2x+cosx
Then,
(du)/dx=2sinxcosx-sinx
Let v=sinx-cos^2x
Then,
(dv)/dx=cosx-2cosx(-sinx)=cosx+2cosxsinx
Rearranging
(dv)/dx=2sinxcosx+cosx
We have,
d/dx(u/v)=(v(du)/dx-u(dv)/dx)/v^2
Substituting
d/dx((sin^2x+cosx)/(sinx-cos^2x))=((sinx-cos^2x)(2sinxcosx-sinx)-(sin^2x+cosx)(2sinxcosx+cosx))/(sinx-cos^2x)^2
Simplifying
((2sin^2xcosx-2sinxcos^3x-sin^2x+cos^2xsinx)-(2sin^3xcosx+2sinxcos^2x+sin^2xcosx+cos^2x))/(sinx-cos^2x)^2
((2sin^2xcosx-2sinxcos^3x-sin^2x+cos^2xsinx-2sin^3xcosx-2sinxcos^2x-sin^2xcosx-cos^2x))/(sinx-cos^2x)^2
(sin^2xcosx-cos^2xsinx-(sin^2x+cos^2x)-2sinx(cos^2x+sin^2x))/(sinx-cos^2x)^2
Knowing that cos^2x+sin^2x=1=sin^2x+cos^2x
(sin^2xcosx-cos^2xsinx-1-2sinx(1))/(sinx-cos^2x)^2
(sinxcosx(sinx-cosx)-1-2sinx)/(sinx-cos^2x)^2
(sinxcosx(sinx-cosx)-(1+2sinx))/(sinx-cos^2x)^2
Thus, the derivative of
(sin^2x+cosx)/(sinx-cos^2x) is
(sinxcosx(sinx-cosx)-(1+2sinx))/(sinx-cos^2x)^2
