The differential of a parametric equation of the type #f(x(t),y(t))# is given by #((dy)/(dt))/((dx)/(dt))#
Here #y(t)=2tcos(pi/2-t/3)# and hence
#(dy)/(dt)=2cos(pi/2-t/3)-2tsin(pi/2-t/3)*(-1/3)#
= #2sin(t/3)+2/3tcos(t/3)# and
as #x(t)=t-tsin(t/2+pi/3)# and hence
#(dx)/(dt)=1-sin(t/2+pi/3)-tcos(t/2+pi/3)*(1/2)#
= #1-sin(t/2+pi/3)-t/2cos(t/2+pi/3)#
= #1-sin(t/2)cos(pi/3)-cos(t/2)sin(pi/3)-t/2[cos(t/2)cos(pi/3)-sin(t/2)sin(pi/3)]#
= #1-1/2sin(t/2)-sqrt3/2cos(t/2)-t/2[1/2cos(t/2)-sqrt3/2sin(t/2)]#
= #1-((2-sqrt3t)/4)sin(t/2)-((2sqrt3+t)/4)cos(t/2)#
and hence
#(df)/(dx)=(2sin(t/3)+2/3tcos(t/3))/(1-((2-sqrt3t)/4)sin(t/2)-((2sqrt3+t)/4)cos(t/2))#
= #(24sin(t/3)+8tcos(t/3))/(12-(6-3sqrt3t)sin(t/2)-(6sqrt3+3t)cos(t/2))#
