[proofplan] The proof uses the transformation law at points fixed by nontrivial elements of $SL_2(\mathbb{Z})$. Define $S := \begin{pmatrix}0 & -1 \\ 1 & 0\end{pmatrix}$ and $T := \begin{pmatrix}1 & 1 \\ 0 & 1\end{pmatrix}$ in $SL_2(\mathbb{Z})$. The matrix $S$ fixes $i$ and contributes the automorphy factor $i$, so the value $f(i)$ must be an eigenvector for multiplication by $i^k$. The matrix $ST = \begin{pmatrix}0 & -1 \\ 1 & 1\end{pmatrix}$ fixes $\rho$ and contributes the automorphy factor $\rho+1 = e^{\pi i/3}$, so the same fixed-point argument forces $f(\rho)$ to vanish unless the corresponding root of unity acts trivially. [/proofplan] [step:Apply the transformation law at the elliptic point $i$] Define the matrix \begin{align*} S := \begin{pmatrix}0 & -1 \\ 1 & 0\end{pmatrix} \in SL_2(\mathbb{Z}). \end{align*} The fractional linear transformation associated to $S$ is the map \begin{align*} S_{\mathbb{H}}: \mathbb{H} &\to \mathbb{H} \\ z &\mapsto -\frac{1}{z}. \end{align*} Since $i^2=-1$, we have \begin{align*} S_{\mathbb{H}}(i) = -\frac{1}{i} = i. \end{align*} For $S$, the lower row is $(c,d)=(1,0)$, so the automorphy factor at $i$ is \begin{align*} ci+d = i. \end{align*} Applying the modular transformation law for $f$ with $\gamma=S$ and $z=i$ gives \begin{align*} f(i) = f(S_{\mathbb{H}}(i)) = i^k f(i). \end{align*} Therefore \begin{align*} (1-i^k)f(i)=0. \end{align*} The powers of $i$ have period $4$, and $i^k=1$ holds exactly when $k \equiv 0 \pmod{4}$. Hence if $k \not\equiv 0 \pmod{4}$, then $1-i^k \neq 0$, and the equation above implies $f(i)=0$. [/step] [step:Apply the transformation law at the elliptic point $\rho$] Define the matrix \begin{align*} A := ST = \begin{pmatrix}0 & -1 \\ 1 & 1\end{pmatrix} \in SL_2(\mathbb{Z}). \end{align*} The fractional linear transformation associated to $A$ is the map \begin{align*} A_{\mathbb{H}}: \mathbb{H} &\to \mathbb{H} \\ z &\mapsto -\frac{1}{z+1}. \end{align*} Since $\rho=e^{2\pi i/3}$ satisfies \begin{align*} \rho^2+\rho+1=0, \end{align*} we have \begin{align*} A_{\mathbb{H}}(\rho) = -\frac{1}{\rho+1} = \rho. \end{align*} For $A$, the lower row is $(c,d)=(1,1)$, so the automorphy factor at $\rho$ is \begin{align*} c\rho+d = \rho+1. \end{align*} Because \begin{align*} \rho+1 = \frac{1}{2}+\frac{\sqrt{3}}{2}i = e^{\pi i/3}, \end{align*} the number $\rho+1$ is a primitive sixth root of unity. Applying the modular transformation law for $f$ with $\gamma=A$ and $z=\rho$ gives \begin{align*} f(\rho) = f(A_{\mathbb{H}}(\rho)) = (\rho+1)^k f(\rho). \end{align*} Therefore \begin{align*} (1-(\rho+1)^k)f(\rho)=0. \end{align*} Since $\rho+1$ is a primitive sixth root of unity, $(\rho+1)^k=1$ holds exactly when $k \equiv 0 \pmod{6}$. Hence if $k \not\equiv 0 \pmod{6}$, then $1-(\rho+1)^k \neq 0$, and the equation above implies $f(\rho)=0$. [/step] [step:Conclude the forced vanishing assertions] The first step proves $f(i)=0$ whenever $k \not\equiv 0 \pmod{4}$, and the second step proves $f(\rho)=0$ whenever $k \not\equiv 0 \pmod{6}$. These are exactly the two asserted forced vanishing statements at the elliptic points $i$ and $\rho$. [/step]