The functions $f$ and $g$ are not continuous and have directional derivative at $(0,0)$ Let $$f:\Bbb{R}^2\to\Bbb{R}$$
$$(x,y)\mapsto f(x,y)=\begin{cases}y^2\log|x|& ,x\neq 0,\\0 &,x=0.\end{cases}$$
and 
$$g:\Bbb{R}^2\to\Bbb{R}$$
$$(x,y)\mapsto f(x,y)=\begin{cases}\frac{x^2 y}{x^2+y^2}& ,(x,y)\neq (0,0),\\\;0 &,x=y=0.\end{cases}$$
I want to show that 


*

*$f$ and $g$ are not continuous at $(0,0).$

*$f$ and $g$ have directional derivatives at $(0,0).$


PART A:
To show that $f$ and $g$ are not continuous, it suffices to use line, i.e. $y=\alpha x,\;\alpha\in \Bbb{R}$. So, I have 
$$f(x,y)=f(x,\alpha x)=\alpha^2 x^2\log|x|$$ which goes to $0$ as $x\to 0$
Also, $$g(x,y)=g(x,\alpha x)=\frac{\alpha x^3 }{x^2+\alpha^2 x^2}=\frac{\alpha x }{1+\alpha^2 \alpha^2}$$ which also goes to $0$ as $x\to 0.$ So, why I'm I not arriving at the desired result? Could my computation be wrong?
PART B:
Let $(x_0,y_0)\in \Bbb{R}^2.$ Then, for $x_0\neq 0,$
$$f((x_0,y_0)+(h_1,h_2))-f(x_0,y_0)$$
$$f(x_0+h_1,y_0+h_2)-f(x_0,y_0)$$
$$=(y_0+h_2)^2\log|x_0+h_1|-y^2_0 \log|x_0|.$$
I got stuck here. So, could anyone please, help-out? I would be glad for any help rendered.
 A: PART A: For $f$ consider the sequences $x_n=\exp(-n^2)$ and $y_n=1/n$. Then $(x_n,y_n)\to(0,0)$, but $f(x_n,y_n)=-1\not\to 0=f(0,0)$.
In contrast to $f$ the function $g$ IS actually continuous at $(0,0)$. To see this note that $|xy|\leq x^2+y^2$ and hence $|g(x,y)|\leq|x|$.
PART B: First recall the definition of directional derivative: A function $f:\mathbb R^2\to\mathbb R$ has a directional derivative at $(x_0,y_0)$ with direction $v=(v_1,v_2)\in\mathbb R^2\backslash\{(0,0)\},||v||=1$, if the limit
\begin{align*}
\lim_{t\to0}\frac{f(x_0+t v_1,y_0+tv_2)-f(x_0,y_0)}{t}
\end{align*} 
exists. (Some authors allow $v$ to be any vector, not necessarily normalized).
To show that $f$ and $g$ have directional derivatives at $(0,0)$ you need to show that the limits $\lim_{t\to 0}\frac{f(tv_1,tv_2)}{t}$ and $\lim_{t\to 0}\frac{g(tv_1,tv_2)}{t}$ exist, where $v=(v_1,v_2)$ is a (normalized) vector in $\mathbb R^2$. For $f$ we obtain 
\begin{align*}
\frac{f(tv_1,tv_2)}{t}=tv_2^2\log|t v_1|\to0\quad\text{ as }t\to0, 
\end{align*}
and for $g$ we have
\begin{align*}
\frac{g(tv_1,tv_2)}{t}=\frac{t^3v_1^2v_2}{t^3(v_1^2+v_2^2)}=v_1^2v_2\quad\text{ for all }t\neq 0.
\end{align*}
Note that since $v=(v_1,v_2)$ has norm 1, we have $1=||v||^2=v_1^2+v_2^2$.
