By derivate rules definition, we have that
Be #y=arctanu#, then #y'=(u')/(1+u^2)#
Thus, in this case let's use chain rule, which states
#(dy)/(dx)=(dy)/(du)(du)/(dx)#, and rename #u=13/x=13x^-1#, and, alike, #v=3/x=3x^-1#.
#(dy)/(dx)=-13x^-2(-13x^-2)/(1+(13/x)^2)-(-3x^-2(-3x^-2)/(1+(3/x)^2))#
#(dy)/(dx)=-13/x^2(-13cancel(x^2))/(cancel(x^2)(x^2+169))+3/x^2(3cancel(x^2))/(cancel(x^2)(x^2+9))#
#(dy)/(dx)=169/(x^2(x^2+169))+9/(x^2(x^2+9))#
Thus, the l.c.d. is #x^2(x^2+169)(x^2+9)#
#(dy)/(dx)=(169(x^2(x^2+9))+9(x^2(x^2+169)))/(x^2(x^2+169)(x^2+9))#
#(dy)/(dx)=(169x^4+1521x^2+9x^4+1521x^2)/(x^2(x^2+169)(x^2+9))#
#(dy)/(dx)=(178x^4+3042x^2)/(x^2(x^2+169)(x^2+9))=(178x^2+3042)/((x^2+169)(x^2+9))#
