On number 4.
In polar coordinates the limit $\lim_{(x,y)\rightarrow (0,0)}f(x,y)$ is equal to
$$\lim_{\rho\rightarrow 0}\frac{\rho^2}{\rho^2}\frac{\cos^2\theta-\sin^2\theta}{\cos^2\theta+\sin^2\theta},$$
whose result is dependent of $\theta$. The function $f$ is not continuous at $(0,0)$.
You can arrive at the same result without polar coordinates and choosing different paths to $(0,0)$, leading to different limit values. For example, try to reach $(0,0)$ along the lines
$$y=x, $$
and
$$y=0.$$
You obtain the limits $\lim_{x\rightarrow 0}f(x,x)=0$ and $\lim_{x\rightarrow (0,0)}f(x,0)=1$. $f$ is not continuous at $(0,0)$, then.
A function $f:\mathbb{R}^2 \to \mathbb{R}$ is differentiable at a point $p$ if there is a linear map $L:\mathbb{R}^2 \to \mathbb{R}$ such that $$\lim_{h \to 0}\frac{\left|f(p+h)-f(p)-L(h)\right|}{|h|} = 0$$ If the function is differentiable, then $L$ is the function $L(x,y) = \dfrac{\partial f }{\partial x}\big|_p x +\dfrac{\partial f}{\partial y}\big|_p y$.
If all you know is that the partial derivatives exist, you do not even know that the function is continuous, let alone is well approximated by a tangent plane. For instance $f(x,y) = 0$ if $x=0$ or $y=0$ and $f(x,y) = 1$ otherwise. The partial derivatives of $f$ at $(0,0)$ are all $0$, but the tangent plane is a really crappy approximation to $f$ off of the coordinate axes.
The example you were looking at is a little harder to visualize, but think about what happens to $f$ on lines other than horizontal and vertical ones.
In other words, morally, for a function to be differentiable at $(a,b)$ we need that $f(a+\Delta x,b+\Delta y) \approx f(a,b) + \frac{\partial f}{\partial x} \Delta x + \frac{\partial f}{\partial y} \Delta y$. This talks about approximating $f$ in all directions around $(a,b)$, whereas existence of the partial derivatives only means you have good approximations in two directions (the coordinate directions).
Best Answer
To show that the function is continuous at the origin, you may use the polar coordinates: putting $x=\rho \cos \theta$ and $ y=\rho \sin \theta$ with $\rho >0$ and $\theta \in [0, 2 \pi)$, you have that $$\sqrt{|xy|}=\sqrt{\rho^2 |\cos \theta \sin \theta|}\le \rho \to 0$$if $\rho \to 0$. About the differentiability, I think that you are right (anyway it is better if you do a background check with the definition).