Complex continuity and differentiability of a piecewise function Let $g(z):\mathbb{C}\rightarrow\mathbb{C}$ with 
$$g(z)=g(x+iy) =  
     \begin{cases} 
       \dfrac{x^2y+ixy^2}{x^2+y^2} & : (x,y)\neq (0,0), \\ 
       0 & : (x,y)=(0,0).
     \end{cases}
$$ 

1 How do I show that $g$ is continuous at $z=0$?
  2 How do I show that $g$ is not complex differentiable at $z=0$? 

What I know:
1 I know that the composition of continuous functions is continuous, but I don't know how to show that the 'subfunctions' of $g$ are continuous. How would I do this with the $\varepsilon$-$\delta$-definition of continuity?
2 I know that $g$ satisfies the Cauchy-Riemann equations at $z=0$, but this is not enough right? My textbook says that $g$ must be also totally differentiable at $z=0$ in the sense of real analysis ($\mathbb{C}=\mathbb{R}^2$). How would I show this?
Edit: $g$ is NOT differentiable at $z=0$.
 A: Outside of the origin, it is convenient to write $g$ in a different manner. The real and imaginary part of the numerator have the common factor $xy$, and taking that out, we can write
$$g(x+iy) = \begin{cases} \dfrac{xy}{x^2+y^2}(x+iy) &, (x,y) \neq (0,0) \\ \qquad 0 &, (x,y) = (0,0). \end{cases}\tag{1}$$
In that form, the answer to both questions is easily found by looking at the factor $\frac{xy}{x^2+y^2}$. The inequality between arithmetic and geometric mean shows
$$\lvert xy\rvert \leqslant \frac{x^2+y^2}{2},$$
and hence $\Bigl\lvert \frac{xy}{x^2+y^2}\Bigr\rvert \leqslant \frac{1}{2}$. This shows
$$\lvert g(x+iy)\rvert \leqslant \frac{1}{2}\lvert x+iy\rvert$$
for $x+iy \neq 0$, from which the continuity of $g$ at $0$ follows easily.
For the question of complex differentiability, $(1)$ lets us write the difference quotient
$$\frac{g(x+iy) - g(0)}{x+iy} = \frac{xy}{x^2+y^2},\tag{2}$$
and one notes that the right hand side has no limit as $(x,y) \to (0,0)$, every value in $\bigl[-\frac{1}{2},\frac{1}{2}\bigr]$ is attained on every circle $\{x+iy : \sqrt{x^2+y^2} = \rho\}$. On the ray $\{ r(\cos \varphi + i\sin\varphi) : r \in (0,+\infty)\}$ the difference quotient has the constant value $\frac{1}{2}\sin (2\varphi)$. Thus $g$ is not complex differentiable at $0$.
This function $g$ shows that it is not sufficient to imply complex differentiability that the partial derivatives exist and satisfy the Cauchy-Riemann equations, (total) real differentiability (Fréchet differentiability) is required.
