For ease of writing, let, #sin^2x=a, and, sin^2y=b...(star_1).#
Then, clearly, #cos^2x=1-a, and, cos^2y=1-b.#
Substituting these, in what is given, we get,
#a^2/b+(1-a)^2/(1-b)=1.#
#:. a^2(1-b)+b(1-a)^2=b(1-b).#
#:. (a^2cancel(-a^2b))+(cancelb-2bacancel(+ba^2))=cancelb-b^2.#
#:. a^2-2ba+b^2=0, i.e., (a-b)^2=0.#
#:. a=b.........................................................................(star_2).#
Now, #sin^4y/sin^2x+cos^4y/cos^2x,#
#=b^2/a+(1-b)^2/(1-a)...................................................[because, (star_1)],#
#=a^2/a+(1-a)^2/(1-a)...................................................[because, (star_2),#
#=a+(1-a)=1.#
# rArr sin^4y/sin^2x+cos^4y/cos^2x=1,# as desired!
Enjoy Maths.!
