Given x+∣y∣=2y ⇒x+y=2y or x−y=2y⇒x=y or x=3y This represent a straight line which passes through origin. Hence, x+∣y∣=2y is continuous at x=0. Now, we check differentiability at x=0 x+∣y∣=2y⇒x+y=2y,y≥0x−y=2y,y<0 Thus, f(x)={x,x/3,y<0y≥0} Now, L.H.D. =h→0−lim−hf(x+h−)fx()=h→0−lim−hx+h−x=−1 R.H.D =h→0+limhf(x+h−)fx()=h→0+limh3x+h−3x=h→0+lim31=31 Since, L.H.D = R.H.D. at x=0 ∴ given function is not differentiable at x=0