Show using the definition of limit that $$\lim _{(x,y)\to(0,1)}\frac{x^2-y^2}{x^2+y^2} = -1$$ and $$\lim_{ (x,y)\to(0,0)}\frac{ (1-\cos(xy))\sin y}{(x^2+y^2) }= 0$$
Definition of limit:
$\lim_{(x,y)\to(a,b)} f(x,y) =L$ if and only if for every $\epsilon >0$ exist $\delta>0$ such that if $((x-a)^2+(y-b)^2)^{1/2}<\delta$ then $|f(x,y)-L|<\epsilon$.
We give more or less full detail for the first problem, and treat the second problem more casually. Let $\epsilon\gt 0$ be given. Note that $$\left|\frac{x^2-y^2}{x^2+y^2}-(-1)\right|=\frac{2x^2}{x^2+y^2}.\tag{1}$$
Suppose that $\delta\le \frac{1}{2}$. We will need to put additional conditions on $\delta$ later.
If $\sqrt{(x-0)^2+(y-1)^2}\lt \delta$, then $(y-1)^2\lt \frac{1}{4}$ and therefore $|y-1|\lt \frac{1}{2}$. It follows that $y\gt \frac{1}{2}$. Thus $y^2\gt \frac{1}{4}$, and therefore $\frac{1}{x^2+y^2}\lt 4$. It follows that $\frac{2x^2}{x^2+y^2}\lt 8x^2$.
We want to make sure that $8x^2\lt \epsilon$. Note that $\sqrt{x^2+(y-1)^2}\gt |x|$. So if $\sqrt{x^2+(y-1)^2}\lt \frac{\sqrt{\epsilon}}{\sqrt{8}}$, then $x^2\lt \frac{\epsilon}{8}$, and therefore $8x^2\lt \epsilon$.
So if we set $\delta=\min(1/2,\sqrt{\epsilon}/\sqrt{8})$, then if $\sqrt{x^2+(y-1)^2}\lt \delta$ then the expression on the left side of (1) will be less than $\epsilon$.
For the second problem, we just mention the basic strategy. We are interested in the absolute value of $\frac{(1-\cos(xy))\sin y}{x^2+y^2}$. Multiply top and bottm by $1+\cos(xy)$, and use the fact that $|\sin t|\le t$. We get that $$\left|\frac{(1-\cos(xy))\sin y}{x^2+y^2}\right|\le \frac{(x^2y^2)|y|}{(|1+\cos(xy)|)(x^2+y^2)}.$$
Now if $\sqrt{x^2+y^2}$ is small enough (and it doesn't have to be very small!) we have $1+\cos(xy)\gt 1$. Also, since $(x-y)^2\ge 0$, we have $\frac{|xy|}{x^2+y^2}\le \frac{1}{2}$ unless $(x,y)=(0,0)$. The $\epsilon$ stuff should now be straightforward. There is a lot of slack. Even if $1-\cos(xy)$ is replaced by its square root, the limit is $0$.