Use epsilon-delta definition of limits to show that
$$\lim_{(x,y) \to (1,2)}(x^2+3x-4y)=-4$$ $$\lim_{(x,y) \to (1,-1)}(x^2+y^2)=2$$
We need to show that $$\forall \epsilon >0( \exists \delta >0( \forall (x,y) \in \mathbb R^2 (0<\sqrt{\left(x-1\right)^{2}+\left(y-2\right)^{2}}<\delta \implies \left|\left(x^2+3x-4y\right)+4\right|<\epsilon)))$$
Taking $\left|x-1\right|,\left|y-2\right|<\delta$ follows : $$\left|\left(x^2+3x-4y\right)+4\right|\le \left|x+4\right|\left|x-1\right|+4\left|y-2\right|< (\delta+5)\left|x-1\right|+4\left|y-2\right|$$
Let $ \delta \le 1$:
$$\left|\left(x^2+3x-4y\right)+4\right|< 6\left|x-1\right|+4\left|y-2\right|$$
So taking $\delta \le \epsilon/10$ proves the claim.
We need to show that $$\forall \epsilon >0( \exists \delta >0( \forall (x,y) \in \mathbb R^2 (0<\sqrt{\left(x-1\right)^{2}+\left(y+1\right)^{2}}<\delta \implies \left|\left(x^2+y^2\right)-2\right|<\epsilon)))$$
$$\left|\left(x^2+y^2\right)-2\right| \le \left|x-1\right|\left|x+1\right| +\left|y-1\right|\left|y+1\right|$$
Taking $\left|x-1\right|,\left|y+1\right|<\delta$ and $\delta \le 1$ follows: $$\left|\left(x^2+y^2\right)-2\right| <3\delta +2\delta$$
So we only need to choose $\delta \le \epsilon/5$.
I want to know if my solution is true,so if you have an alternative solution that would be nice to see that,but please first check mine.
Seems okay; the way you take $\delta$ is not very clear.
I would write the first one as follows. (Similarly for the second one.)
Let $\epsilon>0$. The estimate $$ \begin{align} |x^2+3x-4y-(-4)| &=|x^2+3x-4y-(1^2+3\cdot 1-4\cdot 2)|\\ &\le |x+1|\cdot |x-1|+3|x-1|+4|y-2|\\ &=|x-1+2|\cdot |x-1|+3|x-1|+4|y-2| \end{align} $$ suggests that if you take $$\sqrt{|x-1|^2+|y-2|^2}<\delta\tag{1}$$ then $|x-1|<\delta$ and $|y-1|<\delta$ and thus $$ |x^2+3x-4y-(-4)|\le (\delta+2)\delta+3\delta+4\delta=(\delta+9)\delta\tag{2} $$ So if you take $\delta=\min(1,\epsilon/10)$, then (1) implies $$ |x^2+3x-4y-(-4)|\le (\delta+2)\delta+3\delta+4\delta=(\delta+9)\delta \le 10\delta \le\epsilon. $$