Why is $g(x,y)=\frac{xy}{(x^2+y^2)^{1/2}}$ not differentiable at $(0,0)$? I need to know why the following is not true
Let $g(x,y)=\frac{xy}{(x^2+y^2)^{1/2}}$ when $(x,y)$ not equal to $(0,0)$
and $g(x,y)=0$ when $(x,y)=0$
then  $g(x,y)$ on the $x$ axes$(y=0)$ and the $y$ axes $(x=0)$ is $0$ so the partial derivatives at $(0,0)$ exist and they are zero and since zero is constant function it is continuous hence $g(x,y)$ has continuous partial derivatives and therefore is differentiable at $(0,0)$
 A: Use definition of the differentiability in 2D – Function $g(x,y)$ to be differentiable at $(x_0, y_0)$
$$
\lim_{(x,y) \to (x_0, y_0)} \frac {g(x, y) - g(x_0, y_0)}{\sqrt{\left( x-x_0\right )^2+\left( y-y_0\right )^2}}
$$
should exist. Latter means that the limit should be at least the same when approaching $(x_0, y_0)$ from all directions. Simplest case if you approach $(0,0)$ along rays $y = kx$, so
$$
\lim_{(x,y) \to (x_0, y_0)} \frac {g(x, y) - g(x_0, y_0)}{\sqrt{\left( x-x_0\right )^2+\left( y-y_0\right )^2}} = \lim_{x \to 0} \frac {x^2 k}{x^2 \left( k^2 + 1\right )} = \lim_{x \to 0} \frac k{k^2+1} = \frac k{k^2+1}
$$
which means that limit in fact depends on path you approach $(0,0)$(i.e. on the value of $k$), hence doesn't exists.
A: Theorems:
1.total differentiability $(\Rightarrow)$ directional differentiability (from definition $\Rightarrow$ ) partial differentiability.
2.continuous partial differentiability $\Longrightarrow$ total differentiability.
I believe your main question is why theorem 2 does not apply in your example?
Answer: the partial derivatives are not continuous at $(0,0)$, what you worked out is that the partial derivative exists at $(0,0)$, and $\frac{\partial f}{\partial x}(0,0) = \frac{\partial f}{\partial y}(0,0) = 0 $.

Definition Let $f:\mathbb{R}^n \rightarrow \mathbb{R}$ be a function, and $x_0\in \mathbb{R}^n$. For a $v\in \mathbb{R}^n$, we say that $f$ is directionally differentiable at $x_0$ in the direction $v$ if the limit
$$D_v f(x_0) = \lim_{h\rightarrow 0; h\in\mathbb{R}/\{0\}} \frac{f(x_0 + hv) - f(x_0)}{h}$$
exist. We call $D_v f(x_0)$ the directional derivative of $f$ at $x_0$ in the $v$ direction. If $v = e_i$ is one of the standard basis vectors of $\mathbb{R}^n$, we write $D_v f(x_0)$ as $\frac{\partial f}{\partial x_i}(x_0)$, and refer it as the partial derivative. 
We say $f$ is totally differentiable at $x_0$ if there exists a vector $\triangledown f(x_0) \in \mathbb{R}^n$ with the property that 
$$\lim_{h\rightarrow 0; h\in\mathbb{R}^n/\{0\}} \frac{f(x_0 + h) - f(x_0) - h\cdot\triangledown f(x_0)}{|h|} = 0,$$
we refer to $\triangledown f(x_0)$ (if exists) as the gradient of $f$ at $x_0$.
A: You've already done most of the work. You know that if it is differentiable $Dg = (0, 0)$ and $\lim_{|(x,y)|\rightarrow 0} \frac{g(x,y)}{|(x,y)|} = 0 \iff g(x,y)$ is differentiable. Now, $\forall \delta > 0$, consider $0 < x = y < \delta/2$. Then $\frac{g(x,y)}{|(x,y)|} = \frac{x^2}{2x^2} = 1/2$, which is a contradiction. 
A: For the function to be differentiable at a point, the directional derivative must be a linear function of the direction you're taking the derivative in. In symbols:
$$\lim_{t \rightarrow 0} \frac{f((x,y) + t(a,b)) - f(x,y)}{t}$$
Is a linear function of the vector $(a,b)$. In your case, the directional derivative at $(0,0)$ is:
$$\lim_{t \rightarrow 0} \frac{g((0,0) + t(a,b)) - g(0,0)}{t}$$
$$= \lim_{t \rightarrow 0} \frac{g(ta,tb) - g(0,0)}{t}$$
$$= \lim_{t \rightarrow 0} \frac{\frac{(ta)(tb)}{((ta)^{2} + (tb)^{2})^{1/2}} - 0}{t}$$
$$= \lim_{t\to 0} \frac{t^2 ab}{t^2(a^2 + b^2)^{1/2}} = \frac{ab}{(a^2 + b^2)^{1/2}}$$
This last expression is not a linear function of $(a,b)$, which means $g$ is not differentiable at $(0,0)$. 
You can also show this by computing specific directional derivatives instead of the general case: The directional derivatives in the $x$ and $y$ direction are both zero, for example, as you computed. Now you can compute the directional derivative in the $(1,1)$ direction (ie $x = y$) as:
$$\lim_{x \rightarrow 0} \frac{\frac{x^2}{(x^2 + x^2)^{1/2}}}{x} = \lim_{x \rightarrow 0}\frac{x^2}{x^2\sqrt{2}} = \frac{1}{\sqrt{2}}$$ 
Which is no good, since if the directional derivatives in the $x$ and $y$ direction are zero, by linearity the directional derivative in any direction would have to be zero.
A: Use polar coordinates $x=r\cos(\phi)$ $y=r\sin(\phi)$, your $g(x,y)$ becomes
$$\frac{1}{2}r\sin(2\phi)$$
This function is differantiable.
