I want to prove that
$$\lim_{(x,y)\to(1,0)} \frac{x-y}{x^2+y^2} = 1$$
But I'm out of ideas:
$$\bigg|\frac{x-y}{x^2+y^2} - 1\bigg| = \bigg|\frac{x-y-x^2-y^2}{x^2+y^2} \bigg| \leq \frac{|x||x-1| + |y||y+1|}{|x^2+y^2|} \leq \frac{|x-1|}{|x|} + \frac{|y+1|}{|y|}$$
On
Let $\epsilon > 0$. First notice that $\forall x \in \mathbb{R}$ with $|x-1|<\frac{1}{2}$, we have $|x^2+y^2| \ge x^2+y^2 \ge x^2 > {(\frac{1}{2}})^{2} = \frac{1}{4}$ and $|x| < \frac{3}{2}$. Also, $\forall y \in \mathbb{R}$ with $|y|<\frac{1}{2}$, $|y+1|< \frac{3}{2}$. Take $\delta' = \min{\{\frac{1}{2}, \frac{\epsilon}{12}\}}$ and we have $|{\frac{x-y}{x^2+y^2}-1}| < 4(\frac{3}{2} \cdot \frac{\epsilon}{12} + \frac{\epsilon}{12} \cdot \frac{3}{2}) = \epsilon \, \forall x, y \in \mathbb{R}$ with $|x-1|<\delta'$ and $|y-0|<\delta'$. When $(x,y)\in B_{\delta'}(1,0), |(x-1)^2+y^2| < \delta'$. Hence $|(x-1)^2| < \delta'$ and $|y^2| < \delta'$. Take $\delta = \sqrt{\delta'}$ should suffice.
For the first term use the fact that $\frac {|x-1|} {|x|}<2|x-1|$ as soon as $|x-1|<\frac 1 2$. In the second term you have got into trouble by getting $y^{2}$ in the denominator. Instead it should have $x^{2}$ in that term also. You then have tp make $\frac {y||y+1|} {x^{2}}$ small for which you can use the idea I just mentioned for the first term.