Why is the partial derivative continuous but not differentiable at this point

Question

$$f(x,y)=\begin{cases} 0 & \quad x>1,y>1\\ \sqrt{9-x^2-y^2} & \quad \text{else} \end{cases}$$ There is a point B$(1,1)$ ,and the partial derivatives are continuous at this point
However,$f(x,y)$ is not continuous and not differentiable at this point
If the partial derivative is continuous, function must be differentiable at some point
why is it wrong?Did I get my understanding wrong?

Answer 1

Responding to your comment above in the form of an answer...

The partial derivatives do indeed exist at $(1,1)$, but they are not continuous there. Let's take a look at $f_y$. For $x\le1,y\le1$ we have $$f_y(x,y) = -\frac{y}{\sqrt{9-x^2-y^2}}$$ as you computed above. For $x > 1, y>1$ $f_y(x,y) = 0$. We will now show $f_y$ is discontinous at $(1,1)$. Let $h > 0$, we have $$\lim_{h\to 0^+}|f_y(1+h,1+h) - f(1,1)| = \frac{1}{\sqrt{7}}$$ Thus, $f_y$ is not continous at $(1,1)$. A similar argument shows that $f_x$ is also discontinuous at $(1,1)$. Remember the definition of continuity for multivariable functions – it goes:

$g(x,y)$ is continous at $(x_0,y_0)$ if for all $\varepsilon > 0$, there exists $\delta > 0$ so that for all $(x,y)\neq (x_0,y_0)$ in the open disk centered at $(x_0,y_0)$ of radius $\delta$, we have $$|g(x,y) - g(x_0,y_0)| < \varepsilon.$$