Prove $f$ is not differentiable at $(0,0)$ For 
$$f(x,y)=\begin{cases}
               \frac{x|y|}{\sqrt{x^2+y^2}} & \text{ for }(x,y)\neq (0,0)\\
               0 &  \text{ for } (x,y)=(0,0)
            \end{cases}$$
I'm trying to prove $f$ is not differentiable at $(0,0)$. I showed if $f$ is differentiable at $(0,0),$ then $A=Df_{(0,0)}=0.$ But I don't know how this lead to a contradiction. Anyone has ideas?
 A: If you approach with the paths $(t,0)$ and $(0,t)$, you will conclude that the partial derivatives are $0$.
Zero is particularly convenient because in general, when both partial derivatives are $0$, then any directional derivative must also be $0$ unless $f$ was not differentiable at that point.
Thus in order to show that it is not differentiable at $(0,0)$ it suffices to show a linear path that leads to a different derivative.
One path you may consider is $(t,t)$:
For $t \neq 0$,
$f(t,t)=\frac{t  |{t}|}{\sqrt{2t^2}} =\frac{t  {|t|}}{\sqrt{2} {|t|}} =\frac{t}{\sqrt{2}}$
which leads a derivative of $\frac{1}{\sqrt{2}} \neq 0$.
Note: As mentioned in the comments, it should be clear that the argument is only valid because both partial derivatives are zero. In the general non-zero case, you may use a similar argument by defining a second function $g$ as $f$ minus the linear function that makes both partial derivatives of $g$ equal to zero. $f$ is derivable at a point if and only if $g$ is, thus it would suffice to find a problematic path for $g$ to conclude the non-differentiability of $f$.
A: It's easy. You only have to see that if $f$ is differentiable in $(0,0)$, then $f'(0,0)$ is a linear transformation. So:
$$f'(0,0)\cdot (1,1)=f'(0,0)\cdot((1,0)+(0,1))=f'(0,0)\cdot(1,0)+f'(0,0)\cdot(0,1).$$
And that is a contradiction, because $f'(0,0)\cdot (1,1)\not = f'(0,0)\cdot(1,0)+f'(0,0)\cdot(0,1).$
A: Prove by contradiction. Suppose the contrary that $f$ is differentiable
at $(0,0)$. In particular, partial derivatives $f_{x}(0,0)$ and
$f_{y}(0,0)$ exist.
By direct calculation,
\begin{eqnarray*}
f_{x}(0,0) & = & \lim_{h\rightarrow0}\frac{f(0+h,0)-f(0,0)}{h}\\
 & = & 0
\end{eqnarray*}
because for $h\neq0$, $f(h,0)=0$. Also,
\begin{eqnarray*}
f_{y}(0,0) & = & \lim_{h\rightarrow0}\frac{f(0,0+h)-f(0,0)}{h}\\
 & = & 0.
\end{eqnarray*}
Define the so-called remainder part $R$,  $R:\mathbb{R}^{2}\rightarrow\mathbb{R}$ by
\begin{eqnarray*}
R(x,y) & = & \left[f(x,y)-f(0,0)\right]-\left[f_{x}(0,0)x+f_{y}(0,0)y\right]\\
 & = & f(x,y).
\end{eqnarray*}
That $f$ is differentiable at $(0,0)$ implies that $\lim_{(x,y)\rightarrow(0,0)}\frac{1}{\sqrt{x^{2}+y^{2}}}R(x,y)=0.$
In particular, if we put $x=y=t$ and let $t\rightarrow0+$, then
$(x,y)\rightarrow(0,0)$ (along the straight line $y=x$). Hence,
$\lim_{t\rightarrow0+}\frac{1}{\sqrt{t^{2}+t^{2}}}R(t,t)=\lim_{(x,y)\rightarrow(0,0)}\frac{1}{\sqrt{x^{2}+y^{2}}}R(x,y)=0.$
However, by direct computation,
\begin{eqnarray*}
\lim_{t\rightarrow0+}\frac{1}{\sqrt{t^{2}+t^{2}}}R(t,t) & = & \lim_{t\rightarrow0+}\frac{1}{\sqrt{2}t}f(t,t)\\
 & = & \frac{1}{\sqrt{2}}\lim_{t\rightarrow0+}\frac{1}{t}\cdot\frac{t^{2}}{\sqrt{t^{2}+t^{2}}}\\
 & = & \frac{1}{2},
\end{eqnarray*}
which is a contradiction.
