Proving that the function $xy(-ln(x^2 + y^2))^{1/2}$ is $C^1$ I have to prove that the function $f: A\to\mathbb{R}$, $A = \{(x,y)\in\mathbb{R}^2:x^2+y^2<1\}$
$$f(x,y) = \begin{cases}xy(-\ln{(x^2 + y^2)})^{1/2} & 0<x^2 + y^2 <1 \\ 0 & (x,y) = (0,0) \end{cases}$$
is $C^1 (A)$ and that exist $f_{x,x}$, $f_{yy} \in C(A)$. Then I have to determine if $f \in C^2 (A)$
I started proving that $f\in C^{1} (A)$: for sure we know that $f \in C^1(A\setminus \{0\})$, so we have to show that the partial derivatives are continuous in {$0,0$}. In order to do it we have to study the limits $$\lim_{(x,y)\to (0.0)} -\frac{y[(x^2+y^2)\ln{(x^2+y^2)}+x^2]}{(x^2+y^2)\sqrt{-\ln{(x^2+y^2)}}}$$
and 
$$\lim_{(x,y)\to (0.0)} -\frac{x[(x^2+y^2)\ln{(x^2+y^2)}+y^2]}{(x^2+y^2)\sqrt{-\ln{(x^2+y^2)}}}$$
If this limit is $0$ then $f\in C^1(A)$. I can't find if this limit is $0$. Can you help me with this limit and the other requests? Thank you in advance.
 A: HINT
$-\frac{y((x^2+y^2)\ln{(x^2+y^2)}+x^2)}{(x^2+y^2)\sqrt{-\ln{(x^2+y^2)}}}=
-y\sqrt{-\ln{(x^2+y^2)}}-\frac{yx^2}{(x^2+y^2)\sqrt{-\ln{(x^2+y^2)}}}$
Then take polar coordinates:$$x=r\cos{\theta}$$ $$y=r\sin{\theta}$$
and use that $$\lim_{r \to 0^+}-r\sqrt{-\ln{(r)}}=0$$  $$\lim_{r \to 0}\frac{r}{\sqrt{-\ln{r^2}}}=0$$
Thus the limit is zero.
Use the same argument for the second limit.
A: I'll deal with the first expression. (If this one $\to 0$ then the second one $\to 0$ by symmetry.)
The term $\sqrt{-\ln{(x^2+y^2)}}$ downstairs $\to \infty.$ So it is helping us. Perhaps we can just omit it and the thing will still $\to 0.$ It's worth a try. So consider 
$$\tag 1 \frac{y[(x^2+y^2)\ln{(x^2+y^2)}+x^2]}{x^2+y^2}$$
Slap absolute values on everything and use the facts that $|y|\le (x^2+y^2)^{1/2},$ $x^2 \le x^2 + y^2.$ Let's also write $r= (x^2+y^2)^{1/2}.$ Then the absolute value of $(1)$ is no more that
$$\frac{r[r^2|\ln{r^2}|+r^2]}{r^2} = r|\ln{r^2}| +r.$$
Now you're back in one variable, needing only to show the last expression $\to 0$ as $r\to 0^+.$
