It looks like you also need the dual feasibility condition:
$$\boxed{y\geq 0}$$
The primal problem is:
Minimize: $c^{\top}x$
Subject to: $Ax+b\leq 0; x \in \mathbb{R}^n$
where $c=(3;-2)$ in your specific case.
Given a vector $y\geq 0$, the corresponding unconstrained problem is:
Minimize: $c^{\top}x + y^{\top}(Ax+b)$
Subject to: $x \in \mathbb{R}^n$
You can define the dual function $d(y)$ for $y\geq 0$ as the optimal objective of the unconstrained problem:
\begin{align}
d(y) &= \inf_{x \in \mathbb{R}^n} [c^{\top}x + y^{\top}(Ax+b)]\\
&= \left\{\begin{array}{cc}
y^{\top}b & \mbox{ if $c^{\top} + y^{\top}A=0$} \\
-\infty & \mbox{ else}
\end{array}\right.
\end{align}
where you can argue the answer is indeed $-\infty$ if any component of the vector $c^{\top}+y^{\top}A$ is nonzero (why?). One approach is to maximize the dual function $d(y)$ over all $y\geq 0$, which (to avoid $-\infty$) certainly requires $y$ to be chosen so that:
$$\boxed{c^{\top}+y^{\top}A=0}$$
which is the same as the condition $A^{\top}y=-c=(-3, 2)$.
The feasibility condition is:
$$ \boxed{Ax+b\leq 0} $$
The complementary slackness condition requires $y_i=0$ whenever the corresponding $i$th entry of $Ax + b$ is strictly less than zero, which means we need
$$ \boxed{y^{\top}(Ax+b)=0}$$
Here is a simple way to show that if you have vectors $x,y$ that satisfy the above four boxed conditions (including the equation $y\geq 0$) then $x$ must be an optimal solution to the primal: Since $y$ is a nonnegative vector that satsifies $c^{\top}+y^{\top}A=0$, every $x \in \mathbb{R}^n$ minimizes the corresponding unconstrained problem (for that particular $y$). Thus, the conditions of Theorem III.1 here hold (page 14)
https://ee.usc.edu/stochastic-nets/docs/network-optimization-notes.pdf
which means $x$ solves the primal problem.
