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?

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.
