Let f : [0,1] $ \to $ R be such that f(xy) = f(x).f(y), for all x, y $ \in $ [0, 1], and f(0) $ \ne $ 0. If y = y(x) satiesfies the differential equation, ${{dy} \over {dx}}$ = f(x) with y(0) = 1, then y$\left( {{1 \over 4}} \right)$ + y$\left( {{3 \over 4}} \right)$ is equal to :