I was attempting this problem from munkres: For the function $f:\mathbb{R}^2 \rightarrow \mathbb{R}$ given by $$f(x,y)=\begin{cases} \frac{xy}{x^2+y^2} & \text{ if } (x,y)\neq (0,0)\\ \ \ \ 0 & \ \text{if} \ (x,y)=(0,0)\end{cases}$$
Show that the function is continuous in each variable separately. I was trying to use the $\epsilon-\delta$ definition to solve this one. To show for fixed $y_0$ and $\epsilon>0$, there is a $\delta>0$ with $|x-c|<\delta \implies |f(x,y_0)-f(c,y_0)|<\epsilon$ for an arbitrary $c \in \mathbb{R}$. But I am having trouble finding an upper bound for $|x-c|$ to find a suitable $\delta$.
Here was my effort $|\frac{xy_0}{x^2+y_0^2}-\frac{cy_0}{c^2+y^2}|=|\frac{xy_0(c^2+y_0^2)-cy_0(c^2+y_0^2)}{(x^2+y_0^2)(c^2+y_0^2)}|=|\frac{xy_0c^2+xy_0^3-c^3y_0-cy_{0}^3}{x^2c^2+x^2y_0^2+y_0^2c^2+y_0^4}|$. Now I am trying to get rid of the terms $xy_0c^2$ and $-c^3y_0$ in the numerator so I can have only $y_0^3|x-c|$, but I am struggling to figure out how to do this. I have tried to choose a bound for $|x|$ that depends on $c$, but I have not figured out a good way to simplify the righthand expression, so I can choose a $\delta$ that will successfully complete the proof. I also tried to get rid of the appropriate terms in the denominator, which could provide a nice simplification, however I was unable to find this. Any suggestions?
Edit:I found a hint to this problem through this link and attempted the problem. Is this ok?
$|\frac{xy_0}{x^2+y_0^2}-\frac{cy_0}{c^2+y_0^2}|=|\frac{xy_0(c^2+y_0^2)-cy_0(x^2+y_0^2)}{(x^2+y_0^2)(c^2+y_0^2)}|=|y_0||\frac{x(c^2+y_0^2)-c(c^2+y_0^2)+c(c^2+y_0^2)-c(x^2+y_0^2)}{(x^2+y_0^2)(c^2+y_0^2)}|=|y_0||\frac{(x-c)(c^2+y_0^2)+c(c^2-x^2)}{(x^2+y_0^2)(c^2+y_0^2)}|=|x-c||\frac{y_0(c^2+y_0^2)+cy_0(c+x)}{(x^2+y_0^2)(c^2+y_0^2)}|\leq |x-c|(|\frac{y_0}{x^2+y_0^2}|+ |\frac{x+c}{cy_0}|)\leq $
$|x-c|(|\frac{y_0}{y_0^2}|+|\frac{x+c}{cy_0}|) \leq |x-c|(|\frac{1}{y_0}|+|\frac{x+c}{cy_0}|)$
So now if I let $\epsilon>0$, choose $\text{min} \{1,\frac{\epsilon}{|\frac{1}{y_0}|+|\frac{c+1+|c|}{cy_0}|}\}$ then $|x-c|<\delta \implies |\frac{xy_0}{x^2+y_0^2}-\frac{cy_0}{c^2+y^2}|<\epsilon$.
If I understand your answer correctly, you used the bound $|x+c|\leq|c+1+|c||$, which you cannot do before setting $\delta$ (what happens if $x$ goes to infinity?). However, you can use the following $$ |x+c|\leq |x-c|+2|c| $$ to find a bound that only depends on $|x-c|$ and constants, and which tends to zero as $|x-c|$ does. From there it is only a matter of finding the right $\delta$.