Partial derivative of $f(x,y) = x \arctan\left[\frac{x}{y}\right]$ Can someone help me calculating a partial derivative of the function:
$$f(x,y) =
\begin{cases}
x \arctan\left[\frac{x}{y}\right]  & \text{if } y \neq 0 \\
0 & \text{if } y = 0
\end{cases}$$
To determinate the partial derivative respect to $x$ in $(0,0)$
$$\lim_{h\to 0} \frac{f(x_0+h,y_0)-f(x_0,y_0)}{h}  $$
That becomes 
$$\lim_{h\to 0} \frac{f(0+h,0)-f(0,0)}{h} $$
so
$$\lim_{h\to 0} \frac{h\arctan\left[\frac{h}{0}\right]}{h} $$
However 
$$\lim_{h\to 0} \arctan\left[\frac{h}{0}\right] $$ is indeterminate. The problem is that the the funtion simply substituting becomes indeterminate. However I know that the value of the partial derivative not calculated with the definition is 0. Someone can help me to find the mistake?

$$f_x(0,0)=\lim_{y\to 0}f_x(0,y)=\lim_{y\to 0}\lim_{h\to 0}\frac{f(h,y)-f(0,y)}h$$
$$f_x\left( 0,0 \right) =\lim_{y\rightarrow 0}\lim_{h\rightarrow 0}\arctan \left( \frac{h}{y} \right) 
$$
Because $arctan0=0$, $f_x(0,y)=0$
$$f_x(0,0)=\lim_{y\to 0}0=0$$
This makes sense because $f(0,y)=0$, so its slope must also be $0$
