[proofplan]
The Ricci form of a Kähler metric is exactly the Chern curvature form of the Hermitian metric induced by $\omega$ on the anticanonical bundle $K_X^{-1}$. Therefore positive Ricci form gives a positive Hermitian metric on $K_X^{-1}$, while negative Ricci form gives a positive Hermitian metric on the dual bundle $K_X$. The vanishing case follows because the same curvature form represents the real first Chern class $c_1(X)=c_1(K_X^{-1})$.
[/proofplan]
[step:Compute the Chern curvature of the metric induced on $K_X^{-1}$]
Let $(U_a,z_a)$ be a holomorphic coordinate chart on $X$, where
\begin{align*}
z_a: U_a &\to z_a(U_a)\subseteq \mathbb{C}^n, \\
p &\mapsto (z_{a,1}(p),\dots,z_{a,n}(p)).
\end{align*}
On $U_a$, write the Kähler form as
\begin{align*}
\omega
=
i\sum_{j,k=1}^n g_{a,j\bar{k}}\, dz_{a,j}\wedge d\bar z_{a,k},
\end{align*}
where each coefficient $g_{a,j\bar{k}}:U_a\to \mathbb{C}$ is smooth and the Hermitian matrix
\begin{align*}
G_a(p)=\bigl(g_{a,j\bar{k}}(p)\bigr)_{j,k=1}^n
\end{align*}
is positive definite for every $p\in U_a$.
Define the local holomorphic frame
\begin{align*}
e_a:U_a &\to K_X^{-1},\\
p &\mapsto
\frac{\partial}{\partial z_{a,1}}\bigg|_p\wedge \cdots \wedge
\frac{\partial}{\partial z_{a,n}}\bigg|_p .
\end{align*}
The Hermitian metric on $T^{1,0}X$ determined by $\omega$ induces a Hermitian metric $h_{-1}$ on $K_X^{-1}$, and in the frame $e_a$ its squared norm is
\begin{align*}
|e_a|_{h_{-1}}^2=\det G_a.
\end{align*}
For a Hermitian line bundle with local holomorphic frame $e$ and local squared norm $|e|_h^2$, the Chern curvature form is locally
\begin{align*}
i\Theta_h=-i\,\partial\bar\partial\log |e|_h^2.
\end{align*}
Applying this formula to $(K_X^{-1},h_{-1})$ gives
\begin{align*}
i\Theta_{h_{-1}}
=
-i\,\partial\bar\partial\log \det G_a.
\end{align*}
By the local formula for the Ricci form of a Kähler metric,
\begin{align*}
\operatorname{Ric}(\omega)
=
-i\,\partial\bar\partial\log \det G_a.
\end{align*}
Hence, on every chart $U_a$,
\begin{align*}
i\Theta_{h_{-1}}=\operatorname{Ric}(\omega).
\end{align*}
Since both sides are globally defined real $(1,1)$-forms and the equality holds in every holomorphic coordinate chart, it holds on all of $X$.
[guided]
The point of this step is to identify the Ricci form with the curvature of the naturally induced metric on the anticanonical bundle. We work locally because both the Ricci form and Chern curvature have explicit coordinate formulas.
Let $(U_a,z_a)$ be a holomorphic coordinate chart, with
\begin{align*}
z_a: U_a &\to z_a(U_a)\subseteq \mathbb{C}^n, \\
p &\mapsto (z_{a,1}(p),\dots,z_{a,n}(p)).
\end{align*}
Since $\omega$ is Kähler, it is locally written as
\begin{align*}
\omega
=
i\sum_{j,k=1}^n g_{a,j\bar{k}}\, dz_{a,j}\wedge d\bar z_{a,k},
\end{align*}
where the matrix
\begin{align*}
G_a(p)=\bigl(g_{a,j\bar{k}}(p)\bigr)_{j,k=1}^n
\end{align*}
is Hermitian positive definite at every point $p\in U_a$.
The anticanonical bundle is
\begin{align*}
K_X^{-1}=\Lambda^n T^{1,0}X.
\end{align*}
A natural local holomorphic frame for it is
\begin{align*}
e_a:U_a &\to K_X^{-1},\\
p &\mapsto
\frac{\partial}{\partial z_{a,1}}\bigg|_p\wedge \cdots \wedge
\frac{\partial}{\partial z_{a,n}}\bigg|_p .
\end{align*}
The metric induced by $\omega$ on $T^{1,0}X$ has matrix $G_a$ in this frame of tangent vectors, so the induced determinant metric on $\Lambda^nT^{1,0}X$ satisfies
\begin{align*}
|e_a|_{h_{-1}}^2=\det G_a.
\end{align*}
Now use the Chern curvature formula for a Hermitian holomorphic line bundle. If $e$ is a local holomorphic frame and $|e|_h^2$ is its squared norm, then
\begin{align*}
i\Theta_h=-i\,\partial\bar\partial\log |e|_h^2.
\end{align*}
For the metric $h_{-1}$ on $K_X^{-1}$, this gives
\begin{align*}
i\Theta_{h_{-1}}
=
-i\,\partial\bar\partial\log \det G_a.
\end{align*}
But the Ricci form of the Kähler metric is defined locally by the same expression:
\begin{align*}
\operatorname{Ric}(\omega)
=
-i\,\partial\bar\partial\log \det G_a.
\end{align*}
Thus
\begin{align*}
i\Theta_{h_{-1}}=\operatorname{Ric}(\omega)
\end{align*}
on $U_a$. Since this holds in every holomorphic coordinate chart and both sides are globally defined real $(1,1)$-forms, the equality holds globally on $X$.
[/guided]
[/step]
[step:Deduce positivity of $K_X^{-1}$ from positive Ricci form]
Assume that $\operatorname{Ric}(\omega)$ is positive as a real $(1,1)$-form. From the previous step,
\begin{align*}
i\Theta_{h_{-1}}=\operatorname{Ric}(\omega).
\end{align*}
Therefore the Hermitian metric $h_{-1}$ on $K_X^{-1}$ has positive Chern curvature. By the definition of positivity for a holomorphic line bundle, $K_X^{-1}$ is positive.
[/step]
[step:Deduce vanishing of the real first Chern class from zero Ricci form]
Assume that $\operatorname{Ric}(\omega)=0$. The Chern-Weil representative of the first Chern class of $K_X^{-1}$ is
\begin{align*}
c_1(K_X^{-1})
=
\left[\frac{i\Theta_{h_{-1}}}{2\pi}\right]
\in H^2(X;\mathbb{R}).
\end{align*}
Using $i\Theta_{h_{-1}}=\operatorname{Ric}(\omega)$, we obtain
\begin{align*}
c_1(X)
=
c_1(K_X^{-1})
=
\left[\frac{\operatorname{Ric}(\omega)}{2\pi}\right]
=
0
\end{align*}
in $H^2(X;\mathbb{R})$.
[/step]
[step:Pass to the dual metric to handle negative Ricci form]
Assume that $\operatorname{Ric}(\omega)$ is negative as a real $(1,1)$-form. Let $h_K$ be the Hermitian metric on $K_X$ dual to $h_{-1}$. For dual Hermitian line bundles, the Chern curvature changes sign, so
\begin{align*}
i\Theta_{h_K}
=
-i\Theta_{h_{-1}}.
\end{align*}
Using the curvature identity from the first step gives
\begin{align*}
i\Theta_{h_K}
=
-\operatorname{Ric}(\omega).
\end{align*}
Since $\operatorname{Ric}(\omega)$ is negative, the form $-\operatorname{Ric}(\omega)$ is positive. Hence $h_K$ is a Hermitian metric on $K_X$ with positive Chern curvature, and therefore $K_X$ is positive.
[/step]
