I assume that you are looking for the derivative of #y# with respect to #x# -- that is, for #(dy)/(dx)#.
Use Implicit Differentiation:
#2y^2+2x^2=5#
#d/(dx)(2y^2+2x^2) = d/(dx)(5)#
Now, remember that #y# is some function of #x# that we have left implicit rather than making explicit. (We have not 'solved for y', but we could have.)
So, #d/(dx)(y^2) = 2y (dy)/(dx)# (We're using the Chain Rule here.)
#d/(dx)(2y^2+2x^2) = d/(dx)(5)#
#4y (dy)/(dx) + 4x = 0#
And #dy/dx = -x/y#
(Notice that we could have written #4y (dy)/(dx) + 4x (dx)/(dx) = 0#, but, since #dx/dx = 1#, there's not much point.)
