Derivative exists by first principles but undefined when using chain rule Consider the function $h$ defined by
\begin{align}
h(z,y)=(z^3+y^3)^{\frac{1}{3}}
\end{align}
Then 
\begin{align*}
h_z(0,0)&=\lim_{t\rightarrow 0}\frac{(t^3)^{\frac{1}{3}}}{t}\\
&=1
\end{align*}
When differentiating via the chain rule we have
\begin{align}
h_z(z,y)&=\frac{z^2}{(z^3+y^3)^{\frac{2}{3}}}
\end{align}
This function is not defined at (0,0), but when calculating from first principles we get a well defined answer. What is going on here? 
 A: A function $f$ can have a derivative $f'$ exist at a point, say $x_0$, but the derivative itself is discontinuous at $x_0$.  For example, let $f$ be the function
$$f(x)=
\begin{cases}
x^2\sin(1/x),&x\ne 0\\\\
0,&x=0
\end{cases}
$$
Then, the derivative $f'(x)$ exists everywhere and can be written
$$f'(x)=
\begin{cases}
2x\sin(1/x)-\cos(1/x)&x\ne 0\\\\
0&x=0
\end{cases}
$$
where the derivative at $0$ is found as 
$$f'(0)=\lim_{h\to 0}\frac{h^2\sin(1/h)-0}{h}=0$$
But note that $f'(x)$ is not continuous at $0$ since we have
$$\lim_{x\to 0}\left(2x\sin(1/x)-\cos(1/x)\right)$$
does not even exist (the cosine term oscillates indefinitely)!

Now, the example in the posted question is not different in kind from the aforementioned example.  We have
$$h(z,y)=(z^3+y^3)^{1/3}$$
so that for $(z,y)\ne (0,0)$
$$h_z(z,y)=\frac{z^2}{(z^3+y^3)^{2/3}}$$
and at $(0,0)$ we have
$$h_z(0,0)=\lim_{k\to 0}\frac{h(k,0)-h(0,0)}{k}=\lim_{k\to 0}\frac{k}{k}=1$$
Thus, 
$$h_z(z,y)=
\begin{cases}
\frac{z^2}{(z^3+y^3)^{2/3}}&(y,z)\ne (0,0)\\\\
1&(y,z)=(0,0)
\end{cases}
$$
But, $h_z$ is not continuous at the origin since
$$\lim_{(y,z)\to (0,0)}\frac{z^2}{(z^3+y^3)^{2/3}}$$
is indeterminate (take for example a path along which $y=Cz$).
