= Solution
Let
$$
f(\tau)=\frac{\Delta(\tau)}{\Delta(3\tau)}.
$$
For
$$
\gamma=\begin{pmatrix}a&b\\3c&d\end{pmatrix}\in\Gamma_0(3),
$$
the matrix $\gamma'=\begin{pmatrix}a&3b\\c&d\end{pmatrix}$ lies in $SL_2(\mathbb Z)$ and satisfies $3\gamma\tau=\gamma'(3\tau)$. The weight-twelve transformation law for the <modular discriminant> gives the same factor $(3c\tau+d)^{12}$ in numerator and denominator. Hence $f(\gamma\tau)=f(\tau)$, so $f$ is a weight-zero modular function of level $\Gamma_0(3)$.
The discriminant has no zero in the upper half-plane, so $f$ has neither zeros nor poles there. At infinity,
$$
\Delta(\tau)=q+O(q^2),
\qquad
\Delta(3\tau)=q^3+O(q^6),
$$
and therefore $f=q^{-2}+O(q^{-1})$: it has a pole of order two. The transformation $\Delta(-1/\tau)=\tau^{12}\Delta(\tau)$ gives
$$
f\left(-\frac1{3\tau}\right)
=3^{12}\frac{\Delta(3\tau)}{\Delta(\tau)}
=\frac{3^{12}}{f(\tau)}.
$$
The <Fricke involution> exchanges infinity and zero, so $f$ has a zero of order two at the cusp zero.
Thus the morphism $\phi_f:X_0(3)\to\widehat{\mathbb C}$ has degree two, equal to its total pole order. An isomorphism of compact Riemann surfaces has degree one. Although $X_0(3)$ has genus zero, this particular morphism is therefore not an isomorphism.
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