Proving differentiability without using partial derivatives

Question

Let $f: \mathbb R^{2} \to \mathbb R$,

with

$f(x,y) = \begin{cases} \frac{y^{3}}{\sqrt{x^{2}+y^{2}}} & (x,y) \neq (0,0) \\ 0 & (x,y)=(0,0) \end{cases}$

Show that $f$ is differentiable in $(0,0)$ without calculating the partial derivatives.

Problem:

Using $\lim_{h \to 0}\frac{\vert\vert f(h,h)-f(0,0)-Ah\vert\vert}{\vert\vert h\vert\vert}$=$\lim_{h \to 0}\frac{\vert\vert \frac{h^3}{\sqrt{2h^{2}}}-Ah\vert\vert}{\vert\vert h\vert\vert}=\lim_{h\to0}\frac{\vert\vert\frac{h^2}{\sqrt{2}}-Ah\vert\vert}{\vert\vert h\vert\vert}=\lim_{h\to0}{\vert\vert\frac{h}{\sqrt{2}}-A\vert\vert}$ and that is $\neq 0$ for $h \to 0$ unless I can find $A=0$, however I am unable to use the partial derivatives so I am stuck.

Answer 1

Note that$$\lim_{(x,y)\to(0,0)}\frac{f(x,y)}{\sqrt{x^2+y^2}}=\lim_{(x,y)\to(0,0)}\frac{y^3}{x^2+y^2}=\lim_{(x,y)\to(0,0)}y\frac{y^2}{x^2+y^2}=0.$$Therefore, $f'(0,0)$ is the null function. Indeed, if $\varphi$ is the null function, the previous equality is equivalent to$$\lim_{(x,y)\to(0,0)}\frac{\bigl|f(x,y)-f(0,0)-\varphi(x,y)\bigr|}{\bigl\|(x,y)\bigr\|}=0.$$

Answer 2

The directional derivative is $$df_{_\mathbf{v}}(\mathbf{x})=\lim\limits_{t\to 0}\frac{f(\mathbf{x}+t\mathbf{v})-f(\mathbf{x})}{t}$$

Letting $\mathbf{v}=(h,k)$, you have

$$df_{h,k}(0)=\lim\limits_{t\to 0}\frac{\frac{t^3k^3}{t\sqrt{h^2+k^2}}}{t}=0$$ This is linear, so it remains to show that limit

$$\lim\limits_{(x,y)\to (0,0)}\frac{y^3}{x^2+y^2}=0$$ and this is indeed the case as $y^2\leq x^2+y^2$ and so $\frac{y^3}{x^2+y^2}\leq |y|$.