Use epsilon-delta definition of limits to show that $\lim_{(x,y) \to (1,2)}(3x+2y-1)=6$ and $\lim_{(x,y) \to (0,0)}(x^2+y^2)=0$ Use epsilon-delta definition of limits to show that
$$\lim_{(x,y) \to (1,2)}(3x+2y-1)=6$$
$$\lim_{(x,y) \to (0,0)}(x^2+y^2)=0$$
We need to show that $$\forall \epsilon >0( \exists \delta >0( \forall (x,y) \in \mathbb R^2 (0<\sqrt{\left(x-1\right)^{2}+\left(y-2\right)^{2}}<\delta \implies \left|\left(3x+2y-1\right)-6\right|<\epsilon)))$$
Taking $\left|x-1\right|,\left|y-2\right|<\delta$ follows :
$$\left|\left(3x+2y-1\right)-6\right|\le 3\left|x-2\right|+\left|2y-1\right|<3(\delta+1)+(2\delta+3)$$
$$\left|\left(3x+2y-1\right)-6\right|<5\delta +6$$
Taking $ \delta \le{(\epsilon -6)}/5$ shows the claim.

We need to show that $$\forall \epsilon >0( \exists \delta >0( \forall (x,y) \in \mathbb R^2 (0<\sqrt{x^2+y^2}<\delta \implies \left|x^2+y^2 \right|<\epsilon)))$$
$$\left|x^2+y^2 \right|=x^2+y^2$$
Taking $ \delta \le\sqrt{\epsilon }$ shows the claim.
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.

The latter one is correct, but the former is not. A big hint is that the choice $\delta \le (\varepsilon - 6) / 5$, which is negative(!) when $\varepsilon < 6$. Your choice of $\delta$ should always be positive, so that at least one $(x, y)$ satisfies $0 < \|(x, y) - (x_0, y_0)\| < \delta$.
Instead, consider that
$$|(3x + 2y - 1) - 6| = |3(x - 1) + 2(y - 2)| \le 3|x - 1| + 2|y-2|.$$
So, if we take $\delta = \varepsilon / 5$, then
\begin{align*}
0 < \sqrt{(x - 1)^2 + (y - 2)^2} < \delta &\implies |x - 1|, |y - 2| < \frac{\varepsilon}{5} \\
&\implies |(3x + 2y - 1) - 6| < 3\frac{\varepsilon}{5} + 2\frac{\varepsilon}{5} = \varepsilon.
\end{align*}
