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 $(0,0)$
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.
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.