<s>[INST] Partial Derivative proof For
$$f(x,y) = \begin{cases}
\frac{xy}{x^2+y^2} &\text{if }(x,y)\neq (0,0);\\
0 &\text{if }(x,y)=(0,0).\end{cases}$$
I'm trying to prove that $\frac{\partial f}{\partial x}$, the partial derivative with respect to $x$, exists.
Taking the partial derivative freehand I get
$$\frac{y(x^2+y^2)-xy(2x)}{(x^2+y^2)^2}=  
\frac{y(y^2-x^2)}{(x^2+y^2)^2}.$$
But using the limits I get
$$\lim_{x\to c}\frac{f(x,y)-f(c,y)}{x-c} = y(y^2-c^2).$$
I'm not sure what I'm doing wrong...
 [/INST] I have no idea what you did with the limit. For $y\neq 0$ or $c\neq 0$, we have:
\begin{align*}
\lim{x\to c}\frac{f(x,y)-f(c,y)}{x-c} &= \lim_{x\to c}\frac{\frac{xy}{x^2+y^2} - \frac{cy}{c^2+y^2}}{x-c}\\
&= \lim_{x\to c}\frac{xy(c^2+y^2)-cy(x^2+y^2)}{(c-y)(x^2+y^2)(c^2+y^2)}\\
&= \lim_{x\to c}\frac{(x-c)(y^3-xyc)}{(x-c)(x^2+y^2)(c^2+y^2)}\\
&= \lim{x\to c}\frac{y^3-xyc}{(x^2+y^2)(c^2+y^2)}\\
&= \frac{y(y^2-c^2)}{(c^2+y^2)^2}.
\end{align*}
So, how did you get what you claim you got?
Added. For the partial at $(x,y)=(0,0)$, you have:
$$\lim_{x\to 0}\frac{f(x,0)-f(0,0)}{x-0} = \lim_{x\to 0}\frac{0 - 0}{x} = 0.$$
Since the limits exist, the partial derivative exists at each point.
</s>