Finding partial derivative of $f(x,y)=\frac{xy}{\sqrt{x^2+y^2}}$ I'm studying Mathematical Analysis II for a university course. There is a training exercise that asks me to:

Find the partial derivatives at $(1,0)$ of $f(x,y)$, where:
$f(x,y)=\frac{xy}{\sqrt{x^2+y^2}}$ when $(x,y)\neq (0,0)$ and
$f(x,y) = 0$ when $(x,y)= (0,0)$

So far, I've used the definition as per the other examples. For example, let's start with $f_x(a,b)=g'(a)=\lim_{h\to 0}\frac{f(a+h,b) - f(a,b)}{h}$
Applied to my problem, I've got $f_x(1,0)=\lim_{h\to 0}\frac{f(1+h,0) - f(1,0)}{h}$
And this is where I stall. According to other examples, this is supposed to equal $\frac{0}{h}=0$ , but HOW?
And then I'm supposed to get the limit to +inf?
Am I using the wrong methodology?
 A: Given the function $f$ defined as
$$f(x,y)=
\begin{cases}
\frac{xy}{\sqrt{x^2+y^2}}&,(x,y)\ne 0\\\\
0&,(x,y)=(0,0)
\end{cases}$$
we have
$$\frac{\partial f(x,0)}{\partial x}=\lim_{x\to 0}\frac{f(x+h,0)-f(x.0)}{h}=\lim_{h\to 0}\frac{\frac{(x+h)(0)}{\sqrt{(x+h)^2+o^2}}-\frac{(x)(0)}{\sqrt{(x)^2+o^2}}}{h}=0$$

Note that while both partial derivatives $f_x(0,0)=0$, $f_y(0,0)=1$, the  Derivative of $f$ does not exist at the origin.  We can see this by using the definition of the derivative
$$\lim_{h,k\to 0}\frac{f(0+h,0+k)-f(0,0)-f_x(0,0)h-f_y(0,0)k}{\sqrt{h^2+k^2}}=\lim_{h,k\to 0}\frac{\frac{hk}{\sqrt{h^2+k^2}}-0-0(h)-(1)k}{\sqrt{h^2+k^2}}$$
Since this limit does not exist, $f$ is not differentiable at the origin.
A: $f(1+h,0)=\frac{1+h. 0}{\sqrt{(1+h)^2+0^2}}=0$ similarly $f(1,0)=0$ thus $f_x(1,0)=\lim_{h\to 0}\frac{f(1+h,0) - f(1,0)}{h}=\lim_{h\to 0}\frac{0}{h}=0$
A: At this level you should be allowed to use results from Analysis I (use the tools you already have). Which means that you can calculate the derivatives directly. For example partial derivate of $f(x,y)$ with respect to x at a point $(1,0)$  is the derivate of the function $\phi(x) = f(x,0)$ at the point $x=0$. So we can derive it as if the variable we're not deriving with respect to was a constant:
${\partial\over\partial x}xy(x^2+y^2)^{-1/2} = y(x^2+y^2)^{-1/2} - {1\over2}xy(x^2+y^2)^{-3/2}2x$ 
${\partial\over\partial y}xy(x^2+y^2)^{-1/2} = x(x^2+y^2)^{-1/2} - {1\over2}xy(x^2+y^2)^{-3/2}2y$ 
as for the value at $(0,0)$ it would have no influence for the derivatives at $(1,0)$. Setting $x=1$ and $y=0$ yields:
$f_x'(1,0) = 0(1^2+0^2)^{-1/2} - {1\over2}1\cdot0(1^2+0^2)^{-3/2}2\cdot1 = 0$ 
$f_y'(1,0) = 1(1^2+0^2)^{-1/2} - {1\over2}1\cdot0(1^2+0^2)^{-3/2}2\cdot0 = 1$  
