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Examine the differentiability of $f$: $$f(x,y) = \begin{cases}\displaystyle \frac{x^3}{x^2+y^2} & (x,y) \neq (0,0) \\\\ 0 & (x,y) = (0,0) \end{cases}\:\:. $$

Let's check if $f$ is continuous: let $(x,y) \longrightarrow (0,0)$. Then $$ \left |\frac{x^3}{x^2+y^2}\right| \le |x|\cdot \left|\frac{x^2}{x^2+y^2}\right | \le |x| \longrightarrow 0, $$ so this is ok. Now let's check the continuity of partial derivatives:

\begin{align} \frac{\partial f}{\partial y}(x,y) &= -\frac{2 x^3 y}{\left(x^2+y^2\right)^2}, \\ \frac{\partial f}{\partial x}(x,y) &= \frac{3 x^2}{x^2+y^2}-\frac{2 x^4}{\left(x^2+y^2\right)^2}. \end{align} But I am not sure what should I do now?

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  • $\begingroup$ Those are derivatives for $(x,y) \ne (0,0)$. At the point $(0,0)$ you will have to use the definition of derivative. $\endgroup$ – GEdgar May 28 '19 at 18:28
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Suppose that $f$ is differentiable in $(0,0)$. Then there are a linear map $\mathcal{L}$ and a function $h$, such that

$$f(h_1, h_2) = f(0, 0) + \mathcal{L}(h_1, h_2) + h(h_1, h_2)$$

and

$$\lim_{(h_1,h_2) \to (0,0)} \frac{h(h_1, h_2)}{\sqrt{h_1^2 + h_2^2}} = 0.$$

Notice that $f(0,0) = 0, \ f(h_1, 0) = h_1$ and $f(0, h_2) = 0$, so if there would exist such $\mathcal{L}$ and $h$, it must be $\mathcal{L}(h_1, h_2) = h_1$ for all $(h_1, h_2)$.

Since $f(h_1, h_1) = \frac{1}{2} h_1$, it follows $f(h_1, h_1) - f(0, 0) - \mathcal{L}(h_1, h_1) = - \frac{1}{2}h_1 = h(h_1, h_1)$ and therefore we have

$$\lim_{(h_1,h_1) \to (0,0)} \frac{h(h_1, h_1)}{\sqrt{h_1^2 + h_1^2}} = \lim_{(h_1,h_1) \to (0,0)} - \frac{h_1}{2\sqrt{2h_1^2}} = - \frac{1}{2\sqrt{2}} \neq 0,$$

which is a contradiction. This implies, that $f$ is not differentiable in $(0,0)$.

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Here is a general theorem to keep in mind which is very useful for such questions:

Theorem: Let $V, W$ be normed vector spaces (think of them as $\mathbb{R^n}$ and $\mathbb{R^m}$ if you wish). If $f:V \to W$ is homogeneous, which means for all $t \in \mathbb{R}$ and all $\xi \in V$, \begin{equation} f(t \xi) = t f(\xi) \end{equation} and $f$ is (Frechet) differentiable at $0 \in V$, then $f(\cdot) = df_0(\cdot)$. In particular this says $f$ is linear.

In your case, $f$ is homogeneous, but not linear, hence it can't be differentiable.

For the proof of this theorem, see Loomis and Sternberg's Advanced Calculus, page 148 of the book. There, they discuss this exact example, and show it is not differentiable at the origin.

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