Given that, #f(x-y)=f(x)/f(y).......(star).#
#x=y, &, (star) rArr f(0)=f(y)/f(y)=1.......(1).#
#x=0, y=y, & (star) rArr f(-y)=f(0)/f(y)=1/f(y).........(2).#
#x=x, y=-y, & (star) rArr f(x+y)=f(x)/f(-y)=f(x)f(y)#
#:.," by "(2), f(x+y)=f(x)f(y).............(3).#
Knowing that, #f'(x)=lim_(h to 0) {f(x+h)-f(x)}/h,# we have, by #(3),#
#f'(x)=lim_(h to 0){f(x)f(h)-f(x)}/h=f(x){lim_(h to 0) (f(h)-1)/h}...(ast).#
Now, #f'(0)=p, (ast), & (1) rArr p=lim_(h to 0) (f(h)-1)/h...(4).#
Similarly, #f'(5)=q, (ast), & (4) rArr q=p*f(5)......(5).#
#"Finally, by "(ast), &, (4), f'(-5)=p*f(-5)#
#=p*(1/f(5)),............[because,(2)]#
#=p/(q/p),................[because, (5)]#
#:. f'(-5)=p^2/q,# as Respected Cesareo R., has derived!
Enjoy Maths.!