I assume that you are trying to find #dy/dx#. We think of #y# as a function (or some functions) of #x#, but rather than solving for #y# to make the function explicit, we leave the function implicit.
We will differentiate using the chain rule. So #d/(dx)(y) = (dy)/(dx)# and #d/(dx)(y^2) = 2y (dy)/(dx)# .
For this problem we will also need the product rule, because #x^2y# is 'really'
#x^2 ("some function of x")# like #x^2(g(x))# and similarly for the other term #xy^2#which is 'really' #x*(some function)^2#
#x^2y+xy^2 = 6#. Because these are equal, the derivatives must be equal:
#d/(dx)(x^2y+xy^2) = d/(dx)(6)#
#d/(dx)(x^2y) + d/(dx)(xy^2) = d/(dx)(6)#
#[2xy+x^2 dy/dx] + [1y^2 + x 2y dy/dx] = 0#
#2xy+x^2 dy/dx + y^2 + 2xy dy/dx = 0#
# x^2 dy/dx + 2xy dy/dx = -2xy- y^2 #
# [x^2 + 2xy] dy/dx = -2xy- y^2 #
# dy/dx = (-2xy- y^2) /[x^2 + 2xy] #
