Proving continuity in the origin $\frac{xy}{\sqrt{x^2 + y^2}}$ Let $g: \mathbb{R^2} \to \mathbb{R}$.

How can I prove that  $g$ is continuous in its origin, but not totally differentiable?
If I take 
$$g(\frac{1}{n},\frac{1}{n}) = \frac{1}{n\sqrt{2}} \to 0 \text{ for } n \to \infty $$
Or rather:
$$|x,y| \leq \frac{1}{2} (x^2 + y^2) $$
from which we can follow 
$$|g(x,y)| \leq \frac{1}{2} \sqrt{x^2 + y^2}$$
which proves continuity of $g$ in the origin $(0,0)$.
But how can I show that this function is not total differentiable?
Can I do the following estimation?
$$|g(x,y) - 0| = |y| \cdot \frac{x \cdot y}{\sqrt{x^2 + y^2}} \leq |y| \cdot 1 = |y| $$
From which it follows, that 
$$|y| \leq |\sqrt{x^2 + y^2}| = ||(x,y)|| $$

Note that $x^2+y^2\ge 2|xy|$.  Hence, we have 
$$\left|\frac{xy}{\sqrt{x^2+y^2}}\right| \le \frac{\sqrt{|xy|}}{\sqrt 2}$$
