We define the Lagrangian
$$
\mathcal{L}(x_1, x_2, \lambda_1, \lambda_2) = f_0(x_1, x_2) - \lambda_1 f_1(x_1, x_2) - \lambda_2 f_2(x_1, x_2) - \lambda_3 f_3(x_1, x_2)
$$
Suppose now that all constraints are active. Under Slater's, the 1st order necessary KKT conditions state that
$$
\begin{align}
\partial_{x_1} \mathcal{L} = 0 \\
\partial_{x_2} \mathcal{L} = 0 \\
\partial_{\lambda_1} \mathcal{L} = 0 \\
\partial_{\lambda_2} \mathcal{L} = 0 \\
\partial_{\lambda_3} \mathcal{L} = 0
\end{align}
$$
Now let's see what the solutions to this would look like. These are stated as
\begin{align}
4 x_1 - 15 - 4 x_2 - \lambda_1 + \lambda_3 = 0 \\
8 x_2 -30 - 4 x_1 - \lambda_2 + 2\lambda_3 = 0 \\
x_1 = 0 \\
x_2 = 0 \\
30 - x_1 - 2x_2 = 0
\end{align}
but clearly the three last equations cannot be satisfied. Thus the problem is not minimized with three active constraint.
Suppose now only two constraints are satisfied. Considering first constraint 1 and 2, this yields
\begin{align}
4 x_1 - 15 - 4 x_2 - \lambda_1 = 0 \\
8 x_2 -30 - 4 x_1 - \lambda_2 = 0 \\
x_1 = 0 \\
x_2 = 0 \\
\end{align}
which is a linear system and is satisfied iff $x_1 = 0, x_2 = 0, \lambda_1 = -15, \lambda_2 = -30$. But Lagrange multipliers should be positive.
We continue with constraint 1 and 3
\begin{align}
4 x_1 - 15 - 4 x_2 - \lambda_1 + \lambda_3 = 0 \\
8 x_2 -30 - 4 x_1 + 2\lambda_3 = 0 \\
x_1 = 0 \\
30 - x_1 - 2x_2 = 0
\end{align}
which is satisfied for $x_1 = 0, x_2 = 15, \lambda_3 = 75, \dots$.
$$
f(0, 15) = 450 > f(1, 1) = -43
$$
So this is not our minimizer
We continue with constraint 2 and 3
\begin{align}
4 x_1 - 15 - 4 x_2 + \lambda_3 = 0 \\
8 x_2 -30 - 4 x_1 - \lambda_2 + 2\lambda_3 = 0 \\
x_2 = 0 \\
30 - x_1 - 2x_2 = 0
\end{align}
which is satisfied for $x_1 = 30, x_2 = 0, \lambda_3 = -105$. As lagrange multiplier should be positive this is not our solution either.
Suppose now only one constraint is active,
\begin{align}
4 x_1 - 15 - 4 x_2 - \lambda_1 = 0 \\
8 x_2 -30 - 4 x_1 = 0 \\
x_1 = 0 \\
\end{align}
which yields $x_1 = 0, x_2 = 15/4, \lambda_1 = -75/4$. As before this is not our solution
OR
\begin{align}
4 x_1 - 15 - 4 x_2 = 0 \\
8 x_2 -30 - 4 x_1 - \lambda_2 = 0 \\
x_2 = 0 \\
\end{align}
which yields $x_1 = 15/4, x_2 = 0, \lambda_2 = -45$. As before this is not our solution
OR
\begin{align}
4 x_1 - 15 - 4 x_2 + \lambda_3 = 0 \\
8 x_2 -30 - 4 x_1 + 2\lambda_3 = 0 \\
30 - x_1 - 2x_2 = 0
\end{align}
which yields $x_1 = 12, x_2 = 9, \lambda_3 = 3$. We inspect
$$
f(12, 9) = -270
$$
So clearly a good candidate.
Finally we check that the problem is not solved for all constraints inactive,
\begin{align}
4 x_1 - 15 - 4 x_2 = 0 \\
8 x_2 -30 - 4 x_1 = 0 \\
\end{align}
which yields $x_1 = 15, x_2 = 45/4$. However this solution is not satisfied by the constraints.
To summarize: We considered possible combinations of active constraints and solved the KKT conditions for the (1 + 3 + 3) = 7 different scenario. In all the cases the KKT conditions are linear systems. In the 7 different scenario there are only 2 candidates, $(0, 15)$ and $(12, 9)$. The value attained at $(12, 9)$ is the smallest and we thus conclude that this is the solution to the problem. Wolfram Alpha agrees with me.