= Solution
View $j$ as a <meromorphic function> on $X(\Gamma_0(2))$, using its coordinate $t=j_2$. At the infinity <modular cusp> its <pole> order is one. At the zero <modular cusp>, $j(S\tau)=j(\tau)=q^{-1}+O(1)=q_0^{-2}+O(1)$, so its <pole> order is two. Part (c) identifies these <poles> with $t=\infty$ and $t=0$.
The zero of $j$ on the full modular curve lies at its order-three elliptic point. There are no effective order-three stabilizers in $\Gamma_0(2)$: their lifts have <trace> $\pm1$, whose <characteristic polynomial> modulo two is $X^2+X+1$, whereas an upper triangular <matrix> over $\mathbb F_2$ has both diagonal entries one. Therefore the degree-three covering has exactly one point over that elliptic point, with <analytic ramification index> three. Its $t$-coordinate is $a=j_2(\omega)$, and the zero divisor of the pulled-back $j$ is $3[a]$.
A rational function with these zeros and <poles> must be $C(t-a)^3/t^2$. At the infinity <modular cusp> both $j$ and $t$ have leading coefficient one times $q^{-1}$, giving $C=1$. Hence
$$
\boxed{j(\tau)=\frac{(j_2(\tau)-j_2(\omega))^3}{j_2(\tau)^2}.}
$$
The proof fixes the coefficient and the powers using <modular cusp> <cusp widths> and elliptic ramification, rather than assuming a formula for the level-two coordinate.
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