Continuity of this Piecewise function $f:\mathbb{R}^2\to \mathbb{R}$ I have shown that $f:\mathbb{R}^2\to\mathbb{R}$ given by $f(0,0) = 0$ and $\displaystyle f(x,y)=\frac{x|y|}{\sqrt{x^2+y^2}}$ if $(x,y)\ne (0,0)$ isn't differentiable at $(0,0)$, now I'm trying to show whether it is continuous or not.
My attempt: I must show that $\displaystyle\lim_{(x,y)\to(0,0)}\frac{x|y|}{\sqrt{x^2+y^2}}=f(0,0) = 0$. But $x^2+y^2-2|xy| = (|x|-|y|)^{2}\ge 0$ so $\displaystyle|xy|\le \frac{x^2+y^2}{2}$. So $\displaystyle\left|\frac{x|y|}{\sqrt{x^2+y^2}}\right| =\frac{|xy|}{\sqrt{x^2+y^2}}\le \frac{x^2+y^2}{\sqrt{x^2+y^2}} = (x^2+y^2)^{1/2}$ and once $(x^2+y^2)^{1/2}\to0$ as $(x,y)\to(0,0),$ we would have it.
Is this correct?
 A: Let us consider the directional derivative of $f$ in the direction $v = (a,b)$ at the point $(0,0)$. We have
\begin{align*}
D_{v}f(0,0) & = \lim_{t\to 0}\frac{f((0,0) + t(a,b)) - f(0,0)}{t}\\\\
& = \lim_{t\to 0}\frac{ta|tb|}{t\sqrt{t^{2}a^{2}+t^{2}b^{2}}} = \frac{a|b|}{\sqrt{a^{2}+b^{2}}}
\end{align*}
On the other hand, its partial derivatives are given by
\begin{align*}
\begin{cases}
\displaystyle\frac{\partial f}{\partial x}(0,0) = \lim_{t\to 0}\frac{f((0,0) + t(1,0)) - f(0,0)}{t} = 0\\\\
\displaystyle\frac{\partial f}{\partial y}(0,0) = \lim_{t\to 0}\frac{f((0,0) + t(0,1)) - f(0,0)}{t} = 0
\end{cases}
\end{align*}
Finally, if $f$ was differentiable, we should have that
\begin{align*}
D_{v}f(0,0) = a\frac{\partial f}{\partial x}(0,0) + b\frac{\partial f}{\partial y}(0,0) = a\times 0 + b\times 0 = 0
\end{align*}
which is different from the previous result. Thus $f$ is not differentiable at $(0,0)$.
EDIT
Your approach is fine. Here I propose another way to solve it.
To begin with, notice that $|y| = \sqrt{y^{2}} \leq \sqrt{x^{2} + y^{2}}$. Consequently, one has
\begin{align*}
\frac{|y|}{\sqrt{x^{2}+y^{2}}} \leq 1 \Rightarrow \frac{|xy|}{\sqrt{x^{2}+y^{2}}} \leq |x| \xrightarrow{(x,y)\rightarrow(0,0)} 0
\end{align*}
Hence $f$ is continuous at $(0,0)$.
