We can rewrite $f$ in terms of just $x$ by using the equation of our curve: up to using the equality $y^2=x^3-x$, we get that $f=\frac{(x^3-x)^2+1}{(x^2+1)^3}$. Now things should be pretty straightforwards, since it's easy to tell when this either vanishes or has a pole. There's a full solution under the following spoiler text so you can give it a go yourself using the hint and then check your work.

The numerator vanishes exactly when $x^3-x=\pm i$, and one can check that this gives six distinct possible values for $x$. Since none of these values satisfy $x^3-x$, this means that each possible value gives two distinct $y$-coordinates where it vanishes, so the divisor of zeros is a sum of twelve points (which I hope you'll excuse me for not writing out explicitly). The denominator can be written as $(x+i)^3(x-i)^3$, which vanishes to order 3 at both $x=i$ and $x=-i$, so the divisor of poles is $3(i,\sqrt{-2i})+3(i,-\sqrt{-2i})+3(-i,\sqrt{2i})+3(-i,-\sqrt{2i})$.

In general, it is often the case that local rings of points on an elliptic curve have a nice choice of uniformizer. See for instance here.

The fact that the zeroes and poles of $f \circ [n]$ have the same orders as those of $f$ follows from the facts that: (i) ramification index is multiplicative with respect to composition; and (ii) $[n]$ has ramification index $1$ at each of these points $T'$.

Recall that, given a nonconstant morphism $\varphi: C_1 \to C_2$ of smooth curves and a point $P \in C_1$, the ramification index $e_\varphi(P)$ of $\varphi$ at $P$ is
$$
\DeclareMathOperator{\ord}{ord}
e_\varphi(P) := \ord_P(\varphi^* t)
$$
where $t$ is a local uniformizer at $\varphi(P)$. In the case where $f: C \to \mathbb{P}^1$, $f(P) = 0$, and $t$ is the usual affine coordinate on $\mathbb{P}^1$ at $0$, then $f^* t$ is just $f$ itself, so
$$
e_f(P) = \ord_P(f^* t) = \ord_P(f)
$$
is the order of vanishing of $f$ at $P$.

The two facts above are given in Silverman's *The Arithmetic of Elliptic Curves* Proposition II.2.6, parts (a) and (c):

**Proposition**. Let $\varphi: C_1 \to C_2$ be a nonconstant morphism of smooth curves.

(a) For every $Q \in C_2$,
$$
\sum_{P \in \varphi^{-1}(Q)} e_\varphi(P) = \deg(\varphi).
$$

(c) Given another nonconstant map $\psi: C_2 \to C_3$ of smooth curves, then for all $P \in C_1$, we have
$$
e_{\psi \circ \varphi}(P) = e_\varphi(P) e_\psi(\varphi(P)) \, .
$$

Assuming $n$ is relatively prime to the characteristic of the base field $K$, there are $n^2$ distinct points $T'$ such that $[n]T' = T$ over $\overline{K}$. Since $[n]$ has degree $n^2$, then $e_{[n]}(T') = 1$ for every such $T'$ by part (a). Now applying part (c), we have
\begin{align*}
e_{f \circ [n]}(T') = e_{[n]}(T') e_{f}(T) = 1 \cdot n = n \, .
\end{align*}

## Best Answer

The example problem concerns an elliptic curve $$ y^2 = x^3 + 20x + 20. \tag{1} $$ If $\,x\,$ goes to infinity, then let $\,x=1/t^2\,$ where $\,t\,$ goes to zero and then $$ y = \sqrt{x^3 + 20x + 20} = 1/t^3 + 10t + 10t^3 -50t^5 +O(t^7). \tag{2} $$ Thus, $\,x\,$ has a pole of order $2$ and $\,y\,$ has a pole of order $3$ at infinity. More easily seen by eliminating all but the leading terms of equation $(1)$ to get $\,y^2=x^3\,$ and then $\,y=1/t^3.$

The standard parameterization $$ x = X/Z,\quad y = Y/Z \tag{3} $$ obscures this. A better one often used in cryptography is "Jacobian coordinates" $$ x = X/T^2,\quad y = Y/T^3. \tag{4} $$ Now, given that $$ f = \frac{6y + 71x^2 + 91x + 91}{x^2 + 70x + 11}, \tag{5} $$ the new projectivization in equation $(4)$ gives $$ f = \frac{71X^2 + 6TY + 91T^2 + 91T^4}{X^2+70XT^2+11T^4} . \tag{6} $$ The elliptic curve equation $(1)$ in the new parametrization is $$ Y^2 = X^3+20XT^4+20T^6. \tag{7} $$ The substitution $\,X = Y = 1\,$ and $\,T = 0\,$ satisfies this equation and represents the point at infinity. It also gives $\,f = 71.\,$

In a precise algebraic sense, the poles "cancel out".

Note that the advantage of equation $(3)$ is that it is homogeneous. That is, $\,(X,Y,Z)\,$ represents the same point as $\,(cX,cY,cZ)\,$ for any nonzero $\,c.\,$ For equation $(4)$ the corresponding property is that the tuple $\,(X,Y,T)\,$ represents the same point as $\,(c^2X,c^3Y,cZ)\,$ and this difference must be clearly understood.