I must study the derivability in different directions at point $\overline A$ of $$f(x,y)=\begin{cases} \dfrac{xy-x}{x^2+{(y-1)}^2}&\wedge&(x,y)\neq (0,1)\\0&\wedge&(x,y)=(0,1) \end{cases},\qquad\overline A=(0,1).$$
I did the incremental quotient by definition for a generic versor $\hat r=(u,v)$:
$$\lim_{h\to0}{\dfrac{f((0,1)+h(u,v))-f(0,1)}{h}}= \lim_{h\to0}{\dfrac{f(hu,1+hv)-0}{h}}= \lim_{h\to0}{\dfrac{\frac{hu(1+hv)-hu}{{(hu)}^2+{(hv)}^2}}{h}}= \lim_{h\to0}{\dfrac{hu(1+hv-1)}{h^2\underbrace{\left(u^2+v^2\right)}_{=\; 1}}\cdot\dfrac 1h}= \lim_{h\to0}{\dfrac{huv}{h^2}}= \lim_{h\to0}{\dfrac{uv}h}$$
... but I don't know how to proceed. Is my statement correct? If so, how do I proceed?
Thanks!
I will answer my own question. We have $$\lim_{h\to0}{\frac{uv}{h}}.$$ If we pick $a=0\;\vee\;b=0$ then the quotient is $0$ before calculating the limit, so in those four cases, $ (0,1), (0,-1), (1,0)$ and $(-1,0)$ exists the limit and it's $0$.
Hence for $a$ and $b$ not null does not exist, for $a$ or $b$ null yes and it is $0$.