= Solution
For a divisor $D$ on a smooth projective curve $X$, define
$$
L(D)=\{h\in k(X)^\times:(h)+D\geq0\}\cup\{0\},
\qquad
\ell(D)=\dim_kL(D).
$$
The <Riemann-Roch theorem> states
$$
\boxed{\ell(D)-\ell(K_X-D)=\deg D+1-g},
$$
where $K_X$ is the divisor of a nonzero rational differential, $g$ is the genus of $X$, and $\deg D$ is the sum of the divisor coefficients because the ground field is algebraically closed.
Under the coordinate change $u=1/x$ and $v=y/x^n$, we have
$$
v^2=\frac{y^2}{x^{2n}}
=u^{2n}f(1/u).
$$
Hence
$$
\boxed{g(u)=u^{2n}f(1/u)}.
$$
This is a polynomial, and its value at zero is the nonzero leading coefficient of $f$. Thus the <even-degree hyperelliptic model> has two points $P_+,P_-$ above $u=0$, distinguished by the two values of $v$.
Consider the rational differential
$$
\omega=\frac{dx}{y}.
$$
At a finite branch point $x=a$, square-freeness gives a local parameter $y$, with $x-a$ a nonzero constant times $y^2$ to first order. Hence $dx/y$ is regular and nonzero there. It is also regular and nonzero at finite unramified points.
At infinity,
$$
dx=-u^{-2}du,
\qquad
y=vu^{-n},
$$
so
$$
\omega=-\frac{u^{n-2}}v\,du.
$$
Because $v(P_\pm)\ne0$ and $u$ is a local parameter at each point, $\omega$ has a zero of order $n-2$ at both. Therefore the <canonical divisor of an even-degree hyperelliptic curve> is
$$
\boxed{K_X=(\omega)=(n-2)(P_++P_-)},
$$
and
$$
\boxed{\deg K_X=2n-4}.
$$
Since $\deg K_X=2g-2$, it follows that
$$
2g-2=2n-4,
\qquad
\boxed{g=n-1}.
$$
The function $x$ has a simple pole at each of $P_+$ and $P_-$. Consequently
$$
1,x,x^2,\ldots,x^{n-2}\in L(K_X).
$$
These $n-1$ functions are linearly independent. Riemann-Roch with $D=K_X$ gives $\ell(K_X)=g=n-1$, so
$$
\boxed{L(K_X)=\operatorname{span}_k\{1,x,\ldots,x^{n-2}\}}.
$$
With respect to this basis, the <canonical map of a hyperelliptic curve> is
$$
(x,y)\longmapsto[1:x:x^2:\cdots:x^{n-2}].
$$
It identifies $(x,y)$ with $(x,-y)$ for general $x$, and hence is not an embedding.
If $X$ embedded in $\mathbb P^2$, its smooth plane image would have degree $d\geq4$: degrees one and two have genus zero, and degree three has genus one, whereas $g=n-1\geq2$. But the <canonical map of a smooth plane curve> of degree $d\geq4$ is an embedding by adjunction, contradicting the preceding calculation. Therefore
$$
\boxed{X\text{ cannot be embedded in }\mathbb P^2}.
$$
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