Let $X,Y$ be $\mathbb{P}^{1}_{\mathbb{C}}$, and $f:X\to Y$ is a morphism given by
$$ (x_0:x_1)\mapsto(y_0:y_1)=(x_0^2:x_1^2). $$
I want to prove that
(1)$f_*\mathscr{O}_X$ is locally free sheaf of rank two;
(2)$i:\mathscr{O}_Y\to f_{*}\mathscr{O}_X$ is injective;
(3)The cokernel of $i$ isomorphic to the twisting sheaf $\mathscr{O}_Y(-1)$.
Where $\mathscr{O}_Y(-1)$ is a sheafification of $\mathscr{O}^{'}_Y(-1)$, and $\mathscr{O}^{'}_Y(-1)$ is a presheaf on $Y$ by
$$ \mathscr{O}^{'}_Y(-1)(U) = \{\frac{f}{g}\mid g(P)\neq 0,\forall P \in U\subset Y\}. $$
$f,g$ are homogeneous polyhomials with degree $f =$ degree $g -1$.
I know that the pushforward of locally free sheaf doesn't need to be locally free, but in this special case, this holds true.But I don't know how to prove it.
About (2),I thought of considering the kernel of $i$,but I don't know how to describe the morphism $i$.
Please give me some hints.
The projective line $\mathbb{P}^1$ is covered by the affine lines $D_+(x_0) \cong \mathrm{Spec}(\mathbb{C}[x_1/x_0])$ and $D_+(x_1) \cong \mathrm{Spec}(\mathbb{C}[x_0/x_1])$. Notice that $f$ satisfies $f^{-1}(D_+(x_i))=D_+(x_i)$ for $i=0,1$.
The quasi-coherent sheaf of $\mathcal{O}_Y$-algebras $f_* \mathcal{O}_X$ is given locally by the rings
$(f_* \mathcal{O}_X)(D_+(x_0)) = \mathcal{O}_Y(f^{-1}(D_+(x_0)) = \mathcal{O}_Y(D_+(x_0)) = \mathbb{C}[x_1/x_0]$
$(f_* \mathcal{O}_X)(D_+(x_1)) = \mathcal{O}_Y(f^{-1}(D_+(x_1)) = \mathcal{O}_Y(D_+(x_1)) = \mathbb{C}[x_0/x_1]$
with the algebra structure $f^\# : \mathcal{O}_Y \to f_* \mathcal{O}_X$ given locally by
$f^\#_0 : \mathbb{C}[x_1/x_0] \to \mathbb{C}[x_1/x_0], \quad x_1/x_0 \mapsto x_1^2 / x_0^2$
$f^\#_1 : \mathbb{C}[x_0/x_1] \to \mathbb{C}[x_0/x_1], \quad x_0/x_1 \mapsto x_0^2 / x_1^2$
Now, $f^\#_0$ is free of rank $2$, generated by $\{1,x_1/x_0\}$. You can perhaps see this more clearly by using a homomorphism that is isomorphic to $f^\#_0$ (and thus has the same properties), namely
$g : \mathbb{C}[T^2] \to \mathbb{C}[T], \quad p \mapsto p$
This is free of rank $2$ with basis $\{1,T\}$, which you can see by separating a polynomial into even and odd terms, for example.
Similarly, $f^\#_1$ is free of rank $2$ with basis $\{1,x_0/x_1\}$.
It follows that $f_* \mathcal{O}_X$ is locally free of rank $2$. Since $f^\#_0$ and $f^\#_1$ are injective, the same is true for $f^\#$.
The cokernel of $f^\#_0$ is, as a module, generated by $x_1/x_0$, and similarly the cokernel of $f^\#_1$ is generated by $x_0/x_1$. So the cokernel of $f^\#$ is locally free of rank $1$ and the transition map on the intersection of the two affine lines is $t \mapsto 1/t$, which means that the module is $\mathcal{O}_Y(-1)$.