<s>[INST] Examine differentiability of $f$ 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?
 [/INST] 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 $(0,0)$.
</s>