Existence of partial derivatives & Cauchy-Riemann does not imply differentiability example I learned about the Cauchy-Riemann equations today, and my instructor used the following example to show that differentiability is not guaranteed if the partial derivatives are not continuous.
Let
$$
f(z)=f(x+iy)=
\begin{cases}
\frac{xy(x+iy)}{x^2+y^2},&\quad\mbox{$z\neq 0$}\\
0+0i,&\quad\mbox{$z=0$}.
\end{cases}
$$
Similar to the example here, writing $f(z)=f(x+iy)=u(x,y)+iv(x,y)$, we can show that $u_x(0,0)=0$ and $v_y(0,0)=0$ since the partial derivatives are $0$ on the axes, but $f$ is not continuous and thus not differentiable at $(0,0)$ since the limits are different if we approach $(0,0)$ along the imaginary axis or along $x=y$.
My instructor mentioned that reason why in the above example, differentiability fails even though the partial derivatives exist and satisfy the Cauchy-Riemann equations, is that the partial derivatives are not continuous at $(0,0)$ (there is a similar comment in this document). I wrote out the partial derivatives
\begin{align*}
u_x &= \left(\frac{x^2y}{x^2+y^2}\right)'=\frac{2xy(x^2+y^2)-x^2y(2x)}{(x^2+y^2)^2}=\frac{2xy^3}{(x^2+y^2)^2}\\
v_y &= \left(\frac{xy^2}{x^2+y^2}\right)'=\frac{2yx^3}{(x^2+y^2)^2}.
\end{align*}
E.g., consider $u_x$ - if I approach $(0,0)$ from the real axis, the limit of $u_x$ is $0$; if I approach $(0,0)$ from the line $x=y$, the limit of $u_x$ would be $1/2$. I thought this is why $u_x$ is not continuous at $(0,0)$. But looking at the formulas alone, it seems that the partial derivatives are simply not defined at $(0,0)$ because the denominators would be $0$ (if so, then this would already imply the partial derivatives are not continuous at $(0,0)$ and we wouldn't even need the argument using unequal limits). But we have just shown before that the partial derivatives exist at $(0,0)$. I must be missing something. According to a post here, it may have something to do with the definition of $f$ at $(0,0)$.
Any help is much appreciated!
 A: You're right -- it does have to do with the definition of $f$ at $(0,0)$.
By the definition of partial derivative,
$$ \begin{align} u_x(0,0) &= \lim_{h\to0}\dfrac{u(h,0)-u(0,0)}h \\ &= \lim_{h\to0} \dfrac{\frac{h^20}{h^2+0^2}-0}h \\ &= \lim_{h\to0}\dfrac0h \\ &= 0
\end{align} $$
Notice how we had to use the definition $u(x,y) = \frac{x^2y}{x^2+y^2}$ for $(x,y)=(h,0)$ but the definition $u(x,y) = 0$ for $(x,y) = (0,0)$ because of the piecewise definition of $u(x,y)$. Thus your formula for $u_x$, which (implicitly) used the definition $u(x,y) = \frac{x^2y}{x^2+y^2}$ in both places (that's how the quotient rule is derived from the definition of partial derivative), isn't valid for computing $u_x(0,0)$.
However, your formula is valid for computing $u_x(x,y)$ as long as $(x,y) \neq (0,0)$, since away from $(0,0)$ the function $u$ has a consistent definition. Thus you can use your formula to compute the limits approaching $(0,0)$ (which don't depend on the value at $(0,0)$), as you did. So your proof of discontinuity of $u_x$ is correct.
