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 $1=||v||^2=v_1^2+v_2^2$.
Let $x=r\cos \theta, \quad y=r\sin \theta$
Therefore $$|\frac{xy}{\sqrt{x^2+y^2}}|=r |\cos \theta \sin \theta|\le r=\sqrt{x^2+y^2}\lt \epsilon$$if $x^2\lt\frac{\epsilon^2}{2},\quad$and$\quad y^2\lt \frac{\epsilon^2}{2}$
or if $|x|\lt\frac{\epsilon}{\sqrt2},\quad$and$\quad |y|\lt \frac{\epsilon}{\sqrt2}$
Thus $$|\frac{xy}{\sqrt{x^2+y^2}}-0|\lt \epsilon,\qquad \text{where}\quad |x|\lt\frac{\epsilon}{\sqrt2},\quad and \quad |y|\lt \frac{\epsilon}{\sqrt2}$$
$$\implies \lim_{(x,y)\to (0,0)}\frac{xy}{\sqrt{x^2+y^2}}=0$$
Hence $$\lim_{(x,y)\to (0,0)}f(x,y)=f(0,0)$$and therefore $f(x,y)$ is continuous at $(0,0)$
Again $$f_x(0,0)=\lim_{h \to 0}\frac{f(h,0)-f(0,0)}{h}=0$$and $$f_y(0,0)=\lim_{k \to 0}\frac{f(0,k)-f(0,0)}{k}=0$$
Thus the function $f(x,y)$ possesses partial derivatives at $(0,0)$.
If the given function is differentiable at $(0,0)$, then by definition $$df=f(h,k)-f(0,0)=Ah+Bk+h\phi +k\psi\qquad . . . . . (1)$$ where $\quad A=f_x(0,0)=0;\quad B=f_y(0,0)=0, \quad $and $~\phi,~\psi~$ tends to zero as $\quad (h,k)\to (0,0)$.
So from $(1)$ we have $$\frac{hk}{\sqrt{h^2+k^2}}=h\phi +k\psi$$
Putting $\quad k=mh \quad $and letting $\quad h\to 0,\quad$ we have
$$\frac{m}{\sqrt{1+m^2}}=\lim_{h \to 0}(\phi +m\psi)=0$$which is impossible for arbitraty $~m$.
Hence the function is not differentiable at $(0,0)$.
Best Answer
Hint: $T (h_1 ,h_2) = \nabla f (x_0 , y_0)^T (h_1,h_2) = y_0 h_1 + x_0 h_2 $
Therefore
$$\frac{|f((x_0,y_0)+(h_1,h_2)) - f(x_0,y_0)-T(h_1\space h_2)|}{\|(h_1,h_2)\|}= \frac{ | h_1h_2 | }{\sqrt{h_1^2+h_2^2}} \leq |h_2| $$