Let $C:=V(Y^2Z-X(X-Z)(X-\lambda Z))\subset\mathbb{P}^2$ and define $x:=\frac{X}{Z}$ and $y:=\frac{Y}{Z}$. Letting $z:=\frac{1}{x}$, prove that $L(r(z)_0)\subset k[x,y]$ and $\dim_k L(r(z)_0)=2r$ for all $r\geq 1$.
(this is exercise $8.10$ from Fulton's Algebraic Curves)
Here we use the following notation/terminology:
$\text{div}(f):=\sum_{P\in C}\text{ord}_P(f)P$
$L(D)=\{f\in k(C)\mid \text{div}(f)+D\text{ is effective}\}$
$(z)_0:=\sum_{P\in C, \text{ord}_P(z)>0}\text{ord}_P(z)P$
I've already figured out that (for $P_{\infty}:=(0:1:0)$): $$\text{div}(x)=2(0:0:1)-2P_{\infty}$$ $$\text{div}(y)=(0:0:1)+(1:0:1)+(\lambda:0:1)-2P_{\infty}$$ $$\text{div}(z)=2P_{\infty}-2(0:0:1)$$ $$(z)_0=2P_{\infty}$$ So I have to prove that $L(2rP_{\infty})\subset k[x,y]$ and $\dim_k L(2rP_{\infty})=2r$.
I've noticed that $y^2=x(x-1)(x-\lambda)$, so that $y$ is algebraic over $k(x)$, therefore $k(C)=k(x,y)=k(x)[y]$. So in order to prove that $L(2rP_{\infty})\subset k[x,y]$, I still have to prove $k(x)\cap L(2rP_{\infty})=k[x]$, which I'm not being able to do.
I've also conjectured that a basis for $L(2rP_{\infty})$ is $\{1,x,...,x^r,y,xy,...,x^{r-2}y\}$, which has $2r$ elements, but I couldn't prove in the general case that this is in fact a basis.
1) I shall use the notation $Q_a=(a:0:1)$ so that $$\text{div}(x)=2Q_0-2P_{\infty}$$ $$\text{div}(y)=Q_0+Q_1+Q_\lambda-3P_{\infty}$$ 2) Since $\text{div}(x)=2Q_0-2P_{\infty}$, we have $D:=(z)_0=(1/x)_0=2P_\infty$.
The smooth curve $C$ has degree $3$, hence genus $g=1$.
On the other hand for $r\geq 2, \text {deg}(rD)\geq4\gt 2g-1=1$ , so that Riemann-Roch implies that $$l(rD):=\text {dim}_kL(rD)=1-g+{deg}(rD)=1-1+2r=2r$$
3) If $f\in L(D)$, we have $\text {div}(f)+2rP_\infty\geq 0$, so that $f$ has a pole of order at most $2r$ at infinity but has no pole on the affine part $C_{aff}$ of $C$.
This means that $f$ is regular on $C_{aff}$, i.e. $$f\in k[C_{aff}]=k[x,y]=k[X,Y]/\langle Y^2Z-X(X-Z)(X-\lambda Z\rangle$$ We have thus proved $$L(D)\subset k[x,y]$$