TITLE: On Partial Derivative
QUESTION [8 upvotes]: I am having issues with this mathematical concept, but i couldn't point out where, I've tried rereading thomas and stewart for more than 3 times already but still had no clue. I'll try to explain my thought process and will anyone of you please point out my mistakes?
given a function $f(x,y)=|xy|^{0.5}$, how do we determine if the partial derivative is defined at a point, lets say (0,0)? I've figured out there are 3 methods.
method 1:
$$\frac{d}{dx} |xy|^{0.5} =\frac{(x^2*y^2)^{1/4}}{2x}$$

then at $(0,0)$, it is undefined due to division-by-zero, therefore the P.D is undefined.
method2:
$f(x,0)=0$ 
$$\frac{df}{dx}(0,0) = \frac{d}{dx}f(x,0)\vert_{y=0} = 0$$
then at $(0,0)$, $f_x(0,0) = 0$. (P.D is defined)
method 3:
$$\lim_{h\to 0}\frac{f(0+h,0)-f(0,0)}{h} = 0$$
therefore, $f_x(0,0) = 0$ (P.D is defined)
So what the differences between these method?? Did i apply those methods correctly? If so, why do i have contradicting solutions?
Thanks a lot for the help.

REPLY [5 votes]: Methods 2 and 3 are correct. The partial derivative of $f$ with respect to $x$ at a given point $(x_0,y_0)$ involves the function $g:x\mapsto f(x,y_0)$ and only this function (and moreover, only the values of $g$ in a neighborhood of $x_0$). The values of  $f$ at $(x,y)$ with $y\ne y_0$ are simply not relevant (nor its values at $(x,y)$ for $y=y_0$ and $|x-x_0|\ge\varepsilon$ for any given positive $\varepsilon$).
As pointed out by @Srivatsan Narayanan in a comment, method 3 is essentially the same as method 2. To see why, unroll the definition of the derivative of the single-variable function $x\mapsto f(x,0)$.