[proofplan]
We restrict the entire curve to larger and larger Euclidean discs and compare its pullback metric with the complete hyperbolic metric on each disc. The comparison is exactly the Ahlfors-[Schwarz lemma](/theorems/368): negative holomorphic sectional curvature in the target forces a holomorphic map from a hyperbolic disc to be infinitesimally distance-decreasing up to the factor $\kappa^{-1}$. At a fixed point of $\mathbb{C}$, the hyperbolic metric on the disc of radius $R$ tends to zero as $R\to\infty$, so the pullback of $\omega_Y$ must vanish everywhere. Since $\omega_Y$ is positive definite, $df=0$, and connectedness of $\mathbb{C}$ then forces $f$ to be constant.
[/proofplan]
[step:Compare the restriction of $f$ to each disc with the hyperbolic metric]
For $R>0$, define the open disc
\begin{align*}
\Delta_R:=\{z\in\mathbb{C}: |z|<R\}.
\end{align*}
Let
\begin{align*}
\iota_R:\Delta_R&\to\mathbb{C}\\
z&\mapsto z
\end{align*}
be the inclusion map, and define the restricted holomorphic map
\begin{align*}
f_R:\Delta_R&\to Y\\
z&\mapsto f(\iota_R(z)).
\end{align*}
Equip $\Delta_R$ with the Poincare Kähler form of constant Gaussian curvature $-1$,
\begin{align*}
\omega_R:=\frac{2iR^2}{(R^2-|z|^2)^2}\,dz\wedge d\bar z.
\end{align*}
The Ahlfors-Schwarz lemma applied to $f_R:(\Delta_R,\omega_R)\to (Y,\omega_Y)$ gives
\begin{align*}
f_R^*\omega_Y\leq \frac{1}{\kappa}\,\omega_R
\end{align*}
as Hermitian $(1,1)$-forms on $\Delta_R$.
Here the cited comparison result is the Ahlfors-Schwarz Lemma (citing a result not yet in the wiki: Ahlfors-Schwarz Lemma). Its hypotheses are satisfied because $(\Delta_R,\omega_R)$ is the complete simply connected Riemann surface of constant curvature $-1$, the map $f_R$ is holomorphic, and the target holomorphic sectional curvature is bounded above by $-\kappa$ by hypothesis.
[guided]
Fix a radius $R>0$. We want to use the negative curvature of $Y$, but the map $f$ is defined on all of $\mathbb{C}$, whose flat metric is not the right comparison metric for the Schwarz lemma. The standard device is to restrict $f$ to a hyperbolic disc.
Define
\begin{align*}
\Delta_R:=\{z\in\mathbb{C}: |z|<R\},
\end{align*}
let
\begin{align*}
\iota_R:\Delta_R&\to\mathbb{C}\\
z&\mapsto z
\end{align*}
be the inclusion map, and define
\begin{align*}
f_R:\Delta_R&\to Y\\
z&\mapsto f(\iota_R(z)).
\end{align*}
Since $f$ and $\iota_R$ are holomorphic, their composition $f_R$ is holomorphic.
Now put on $\Delta_R$ the Poincare Kähler form
\begin{align*}
\omega_R:=\frac{2iR^2}{(R^2-|z|^2)^2}\,dz\wedge d\bar z.
\end{align*}
This is the complete hyperbolic metric on $\Delta_R$ with Gaussian curvature $-1$. The Ahlfors-Schwarz Lemma says that if a holomorphic map goes from a complete hyperbolic disc of curvature $-1$ into a Kähler manifold whose holomorphic sectional curvature is at most $-\kappa$, then its pullback metric is bounded above by $\kappa^{-1}$ times the hyperbolic metric on the source. Applying that lemma to
\begin{align*}
f_R:(\Delta_R,\omega_R)\to (Y,\omega_Y)
\end{align*}
gives
\begin{align*}
f_R^*\omega_Y\leq \frac{1}{\kappa}\,\omega_R.
\end{align*}
The hypotheses of the comparison theorem are exactly the ones available here: $f_R$ is holomorphic, $\omega_R$ is the complete curvature $-1$ metric on $\Delta_R$, and the holomorphic sectional curvature of $\omega_Y$ is bounded above by $-\kappa$ at every point and in every nonzero complex tangent direction.
[/guided]
[/step]
[step:Let the radius tend to infinity to force the pullback metric to vanish]
Fix $z_0\in\mathbb{C}$. For every $R>|z_0|$, the point $z_0$ belongs to $\Delta_R$, and the preceding estimate gives
\begin{align*}
(f^*\omega_Y)_{z_0}=(f_R^*\omega_Y)_{z_0}\leq \frac{1}{\kappa}\frac{2iR^2}{(R^2-|z_0|^2)^2}\,dz\wedge d\bar z\bigg|_{z_0}.
\end{align*}
The scalar coefficient on the right satisfies
\begin{align*}
\frac{2R^2}{\kappa(R^2-|z_0|^2)^2}\to 0
\end{align*}
as $R\to\infty$. Since $f^*\omega_Y$ is a semipositive $(1,1)$-form, it follows that
\begin{align*}
(f^*\omega_Y)_{z_0}=0.
\end{align*}
Because $z_0\in\mathbb{C}$ was arbitrary,
\begin{align*}
f^*\omega_Y=0
\end{align*}
on $\mathbb{C}$.
[guided]
Now we use the fact that the whole complex plane is exhausted by larger and larger discs. Fix a point $z_0\in\mathbb{C}$. Whenever $R>|z_0|$, the point $z_0$ lies inside $\Delta_R$, so the estimate from the previous step can be evaluated at $z_0$:
\begin{align*}
(f^*\omega_Y)_{z_0}=(f_R^*\omega_Y)_{z_0}\leq \frac{1}{\kappa}(\omega_R)_{z_0}.
\end{align*}
Substituting the explicit formula for $\omega_R$ gives
\begin{align*}
(f^*\omega_Y)_{z_0}\leq \frac{1}{\kappa}\frac{2iR^2}{(R^2-|z_0|^2)^2}\,dz\wedge d\bar z\bigg|_{z_0}.
\end{align*}
The important point is that $z_0$ is fixed while $R$ tends to infinity. The coefficient of the form on the right is
\begin{align*}
\frac{2R^2}{\kappa(R^2-|z_0|^2)^2}.
\end{align*}
Dividing numerator and denominator by $R^4$ shows that this coefficient equals
\begin{align*}
\frac{2}{\kappa R^2\left(1-\frac{|z_0|^2}{R^2}\right)^2},
\end{align*}
which tends to $0$ as $R\to\infty$.
The form $f^*\omega_Y$ is semipositive because it is the pullback of the positive Kähler form $\omega_Y$ by a holomorphic map. A semipositive Hermitian form bounded above by arbitrarily small positive multiples of the Euclidean form must be zero. Hence
\begin{align*}
(f^*\omega_Y)_{z_0}=0.
\end{align*}
Since $z_0$ was arbitrary, we obtain
\begin{align*}
f^*\omega_Y=0
\end{align*}
on all of $\mathbb{C}$.
[/guided]
[/step]
[step:Use positivity of the Kähler form to show that $df$ vanishes]
Let $J_{\mathbb{C}}:T\mathbb{C}\to T\mathbb{C}$ denote the standard complex structure on $\mathbb{C}$, and let $J_Y:TY\to TY$ denote the complex structure of the Kähler manifold $Y$. Let $z_0\in\mathbb{C}$ and let $v\in T_{z_0}\mathbb{C}$. Since $f^*\omega_Y=0$,
\begin{align*}
0=(f^*\omega_Y)_{z_0}(v,J_{\mathbb{C}}v)=\omega_Y{}_{f(z_0)}\bigl(df_{z_0}(v),J_Ydf_{z_0}(v)\bigr).
\end{align*}
The Kähler metric associated to $\omega_Y$ is positive definite, so
\begin{align*}
\omega_Y{}_{f(z_0)}\bigl(\eta,J_Y\eta\bigr)>0
\end{align*}
for every nonzero $\eta\in T_{f(z_0)}Y$. Therefore $df_{z_0}(v)=0$. Since $z_0$ and $v$ were arbitrary,
\begin{align*}
df=0
\end{align*}
on $\mathbb{C}$.
[/step]
[step:Conclude that the holomorphic map is constant]
Because $df=0$, the map $f$ is locally constant on $\mathbb{C}$. The domain $\mathbb{C}$ is connected, so a locally constant map from $\mathbb{C}$ to $Y$ is constant. Thus every holomorphic map
\begin{align*}
f:\mathbb{C}\to Y
\end{align*}
is constant, as claimed.
[/step]
