Consider f(x,y)={x2yx2+y2if (x,y)≠(0,0)0if (x,y)=(0,0)f(x, y) = \begin{cases} \frac{x^2 y}{x^2 + y^2} & \text{if } (x, y) \neq (0, 0) \\ 0 & \text{if } (x, y) = (0, 0) \end{cases}f(x,y)={x2+y2x2y0if (x,y)=(0,0)if (x,y)=(0,0). Is fff differentiable at (0,0)(0, 0)(0,0)?
No, fff is not continuous at (0,0)(0, 0)(0,0)
Yes, fff is differentiable with ∇f(0,0)=(0,0)\nabla f(0,0) = (0, 0)∇f(0,0)=(0,0)
No, fx(0,0)f_x(0, 0)fx(0,0) and fy(0,0)f_y(0, 0)fy(0,0) do not exist
Yes, but ∇f(0,0)\nabla f(0, 0)∇f(0,0) is undefined