Does $u,v$ possess continuous partials at $z_0?$ I know from Cauchy-Riemann equation that if $f=u+iv$ is differentiable at a point $z_0$ then the 1st order partials of $u,v$ exist at $z_0$ and satisfy C-R equation there. My question is in such a case does $u,v$ possess continuous partials at $z_0?$
 A: Not necessarily. Consider the following (admittedly contrived) example:
Let 
$$f(z) = \left\{\begin{array}{ll}|z|^2\sin\left(\frac{1}{|z|}\right)(1+i)&\text{if $z\neq 0$}\\0&\text{if $z= 0$}\end{array}\right.$$
Then, we observe the function is differentiable at the origin, since
$$
\lim_{z\to 0}\frac{f(z)-f(0)}{z} =\lim_{z\to 0}\frac{|z|^2\sin\left(\frac{1}{|z|}\right)(1+i)-0}{z} = 0
$$
If you want to verify this limit, you can show its absolute value goes to zero. As the CR equations imply, the partials for $u$ and $v$ exist at the origin. In this case, $u$ and $v$ are
$$
u(x,y) = v(x,y) = \left\{\begin{array}{ll}(x^2+y^2)\sin\left(\frac{1}{x^2+y^2}\right)&\text{if $z\neq 0$}\\0&\text{if $z= 0$}\end{array}\right.$$
so their partials at the origin are
$$
u_x = v_x = u_y = v_y = \lim_{h\to0}\frac{h^2\sin h^{-1} - 0}{h} = \lim_{h\to0}h\sin h^{-1}=0
$$
But one can verify that $u=v$ does not have continuous partials at the origin (http://mathinsight.org/differentiable_function_discontinuous_partial_derivatives).
A: Alec's answer is of course correct.
One comment. If $u,v$ have partial derivatives in an open set $U$ and satisfy the Cauchy-Riemann equation in $U$, then their partial derivatives are continuous and $f=u+iv$ is holomorphic in $U$, as a consequence of Looman-Menchoff Theorem.
